152519
A battery of emf $8 \mathrm{~V}$ and internal resistance 0.5 $\Omega$ is being charged by a $120 \mathrm{~V}$ d.c. supply using a series resistor of $15.5 \Omega$. The terminal voltage of $8 \mathrm{~V}$ battery during charging is
1 $11.5 \mathrm{~V}$
2 $112 \mathrm{~V}$
3 $14 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, emf of the storage battery $(\mathrm{E})=8 \mathrm{~V}$ Internal resistance of the battery $(\mathrm{r})=0.5 \Omega$ DC supply voltage $(\mathrm{V})=120 \mathrm{~V}$ Resistance of the resistor $(\mathrm{R})=15.5 \Omega$ $\mathrm{R}$ is connected to the storage battery in series. Effective voltage in the circuit $\left(V^{\prime}\right)=V-E$ $=120-8$ $=112 \mathrm{~V}$ According to Ohm's law, $\mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}+\mathrm{r}}=\frac{112}{15.5+0.5}=\frac{112}{10}=7 \mathrm{~A}$ Voltage across resistance $\mathrm{V} =\mathrm{IR}$ $=7 \times 15.5=108.5 \mathrm{~V}$ DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $\mathrm{R}$ Terminal voltage of battery $=120-108.5$ $=11.5 \mathrm{~V}$
AP EAMCET-28.04.2017
Current Electricity
152520
A circuit is shown in the figure. The e.m.f. of the battery is
1 $4 \mathrm{~V}$
2 $6 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
C $2 \Omega$ and $1 \Omega$ resistance are in series. $2 \Omega$ resistor through which $1 \mathrm{~A}$ current is flowing, we get- $\mathrm{V}_{\mathrm{CO}}=1 \times 2=2 \mathrm{~V}$ $\therefore \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CO}}}{3}=\frac{2}{3} \mathrm{~A}$ Applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{B}}=1+\mathrm{I}_{\mathrm{C}}=1+\frac{2}{3}=\frac{5}{3} \mathrm{~A}$ $\because \mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{B}} \times 2=\frac{5}{3} \times 2=\frac{10}{3} \mathrm{~V}$ $\therefore \quad \mathrm{V}_{\mathrm{BO}}=\mathrm{V}_{\mathrm{BC}}+\mathrm{V}_{\mathrm{CO}}=\frac{10}{3}+2=\frac{16}{3} \mathrm{~V}$ $\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{V}_{\mathrm{BO}}}{6}=\frac{16 / 3}{6}=\frac{8}{9} \mathrm{~A}$ Again applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}$ $=\frac{5}{3}+\frac{8}{9}=\frac{23}{9} \mathrm{~A}$ $\mathrm{~V}_{\mathrm{AB}}=\mathrm{I}_{\mathrm{A}} \times 2=\frac{23}{9} \times 2=\frac{46}{9} \mathrm{~V}$ Therefore, $\mathrm{V}_{\mathrm{AO}}=\mathrm{V}_{\mathrm{AB}}+\mathrm{V}_{\mathrm{BO}}=\frac{46}{9}+\frac{16}{3}=\frac{94}{9} \mathrm{~V}$ Voltage drop across the battery is the same as the voltage drop between $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$. The emf of the battery $\mathrm{E}=\mathrm{V}_{\mathrm{AO}}=\frac{94}{9}=10.44 \mathrm{~V} \approx 12 \mathrm{~V}$
AMU-2017
Current Electricity
152521
Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is
1 $1: 1$
2 $1: 2$
3 $1: 3$
4 $1: 4$
Explanation:
A Let, the emf of a cell $=\mathrm{E}$ Internal Resistance $=\mathrm{r}$ then in series, $\mathrm{P}_{1}=\left(\frac{2 \mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)^{2} \times 2 \mathrm{r}=2 \mathrm{E}^{2} \mathrm{r}$ In parallel, $\mathrm{P}_{2}=\left(\frac{\mathrm{E}}{\mathrm{r} / 2}\right)^{2} \times \mathrm{r} / 2=2 \mathrm{E}^{2} \mathrm{r}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2 \mathrm{E}^{2} \mathrm{r}}{2 \mathrm{E}^{2} \mathrm{r}}$ $\mathrm{P}_{1}: \mathrm{P}_{2}=1: 1$
AMU-2017
Current Electricity
152523
In the circuit shown below emf of each battery is $5 \mathrm{~V}$; and has an internal resistance of $1.0 \Omega$. The current in the circuit $\left(\mathrm{I}_{0}\right)$ and the reading in an ideal voltmeter $(\mathrm{V})$ are:
1 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=0$
2 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=0$
3 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=1$
4 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=1$
Explanation:
B Given, Internal resistance (r) $=1 \Omega$, And emf $(\mathrm{E})=5 \mathrm{~V}$ Let, $I$ is the current flow in the circuit Net emf of the circuit $=8 \times 5 \mathrm{~V}=40 \mathrm{~V}$ and Net resistance in the circuit $=8 \times 1 \Omega=8 \Omega$ then, current in the circuit $\mathrm{I}=\frac{40}{8}=5 \mathrm{~A}$ And voltmeter reading $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=5-5 \times 1$ $\mathrm{~V}=0$
TS EAMCET(Medical)-2017
Current Electricity
152524
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is connected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is
1 $6 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
D Given, emf of each batteries $(E)=20 \mathrm{~V}$ External Resistance $(\mathrm{R})=4 \Omega$ Voltage drop across $\mathrm{AB}$ and $\mathrm{CD}$ are same i.e. $20 \mathrm{~V}$ Hence, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20}{4}=5 \mathrm{~A}$
152519
A battery of emf $8 \mathrm{~V}$ and internal resistance 0.5 $\Omega$ is being charged by a $120 \mathrm{~V}$ d.c. supply using a series resistor of $15.5 \Omega$. The terminal voltage of $8 \mathrm{~V}$ battery during charging is
1 $11.5 \mathrm{~V}$
2 $112 \mathrm{~V}$
3 $14 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, emf of the storage battery $(\mathrm{E})=8 \mathrm{~V}$ Internal resistance of the battery $(\mathrm{r})=0.5 \Omega$ DC supply voltage $(\mathrm{V})=120 \mathrm{~V}$ Resistance of the resistor $(\mathrm{R})=15.5 \Omega$ $\mathrm{R}$ is connected to the storage battery in series. Effective voltage in the circuit $\left(V^{\prime}\right)=V-E$ $=120-8$ $=112 \mathrm{~V}$ According to Ohm's law, $\mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}+\mathrm{r}}=\frac{112}{15.5+0.5}=\frac{112}{10}=7 \mathrm{~A}$ Voltage across resistance $\mathrm{V} =\mathrm{IR}$ $=7 \times 15.5=108.5 \mathrm{~V}$ DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $\mathrm{R}$ Terminal voltage of battery $=120-108.5$ $=11.5 \mathrm{~V}$
AP EAMCET-28.04.2017
Current Electricity
152520
A circuit is shown in the figure. The e.m.f. of the battery is
1 $4 \mathrm{~V}$
2 $6 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
C $2 \Omega$ and $1 \Omega$ resistance are in series. $2 \Omega$ resistor through which $1 \mathrm{~A}$ current is flowing, we get- $\mathrm{V}_{\mathrm{CO}}=1 \times 2=2 \mathrm{~V}$ $\therefore \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CO}}}{3}=\frac{2}{3} \mathrm{~A}$ Applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{B}}=1+\mathrm{I}_{\mathrm{C}}=1+\frac{2}{3}=\frac{5}{3} \mathrm{~A}$ $\because \mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{B}} \times 2=\frac{5}{3} \times 2=\frac{10}{3} \mathrm{~V}$ $\therefore \quad \mathrm{V}_{\mathrm{BO}}=\mathrm{V}_{\mathrm{BC}}+\mathrm{V}_{\mathrm{CO}}=\frac{10}{3}+2=\frac{16}{3} \mathrm{~V}$ $\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{V}_{\mathrm{BO}}}{6}=\frac{16 / 3}{6}=\frac{8}{9} \mathrm{~A}$ Again applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}$ $=\frac{5}{3}+\frac{8}{9}=\frac{23}{9} \mathrm{~A}$ $\mathrm{~V}_{\mathrm{AB}}=\mathrm{I}_{\mathrm{A}} \times 2=\frac{23}{9} \times 2=\frac{46}{9} \mathrm{~V}$ Therefore, $\mathrm{V}_{\mathrm{AO}}=\mathrm{V}_{\mathrm{AB}}+\mathrm{V}_{\mathrm{BO}}=\frac{46}{9}+\frac{16}{3}=\frac{94}{9} \mathrm{~V}$ Voltage drop across the battery is the same as the voltage drop between $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$. The emf of the battery $\mathrm{E}=\mathrm{V}_{\mathrm{AO}}=\frac{94}{9}=10.44 \mathrm{~V} \approx 12 \mathrm{~V}$
AMU-2017
Current Electricity
152521
Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is
1 $1: 1$
2 $1: 2$
3 $1: 3$
4 $1: 4$
Explanation:
A Let, the emf of a cell $=\mathrm{E}$ Internal Resistance $=\mathrm{r}$ then in series, $\mathrm{P}_{1}=\left(\frac{2 \mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)^{2} \times 2 \mathrm{r}=2 \mathrm{E}^{2} \mathrm{r}$ In parallel, $\mathrm{P}_{2}=\left(\frac{\mathrm{E}}{\mathrm{r} / 2}\right)^{2} \times \mathrm{r} / 2=2 \mathrm{E}^{2} \mathrm{r}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2 \mathrm{E}^{2} \mathrm{r}}{2 \mathrm{E}^{2} \mathrm{r}}$ $\mathrm{P}_{1}: \mathrm{P}_{2}=1: 1$
AMU-2017
Current Electricity
152523
In the circuit shown below emf of each battery is $5 \mathrm{~V}$; and has an internal resistance of $1.0 \Omega$. The current in the circuit $\left(\mathrm{I}_{0}\right)$ and the reading in an ideal voltmeter $(\mathrm{V})$ are:
1 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=0$
2 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=0$
3 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=1$
4 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=1$
Explanation:
B Given, Internal resistance (r) $=1 \Omega$, And emf $(\mathrm{E})=5 \mathrm{~V}$ Let, $I$ is the current flow in the circuit Net emf of the circuit $=8 \times 5 \mathrm{~V}=40 \mathrm{~V}$ and Net resistance in the circuit $=8 \times 1 \Omega=8 \Omega$ then, current in the circuit $\mathrm{I}=\frac{40}{8}=5 \mathrm{~A}$ And voltmeter reading $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=5-5 \times 1$ $\mathrm{~V}=0$
TS EAMCET(Medical)-2017
Current Electricity
152524
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is connected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is
1 $6 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
D Given, emf of each batteries $(E)=20 \mathrm{~V}$ External Resistance $(\mathrm{R})=4 \Omega$ Voltage drop across $\mathrm{AB}$ and $\mathrm{CD}$ are same i.e. $20 \mathrm{~V}$ Hence, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20}{4}=5 \mathrm{~A}$
152519
A battery of emf $8 \mathrm{~V}$ and internal resistance 0.5 $\Omega$ is being charged by a $120 \mathrm{~V}$ d.c. supply using a series resistor of $15.5 \Omega$. The terminal voltage of $8 \mathrm{~V}$ battery during charging is
1 $11.5 \mathrm{~V}$
2 $112 \mathrm{~V}$
3 $14 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, emf of the storage battery $(\mathrm{E})=8 \mathrm{~V}$ Internal resistance of the battery $(\mathrm{r})=0.5 \Omega$ DC supply voltage $(\mathrm{V})=120 \mathrm{~V}$ Resistance of the resistor $(\mathrm{R})=15.5 \Omega$ $\mathrm{R}$ is connected to the storage battery in series. Effective voltage in the circuit $\left(V^{\prime}\right)=V-E$ $=120-8$ $=112 \mathrm{~V}$ According to Ohm's law, $\mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}+\mathrm{r}}=\frac{112}{15.5+0.5}=\frac{112}{10}=7 \mathrm{~A}$ Voltage across resistance $\mathrm{V} =\mathrm{IR}$ $=7 \times 15.5=108.5 \mathrm{~V}$ DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $\mathrm{R}$ Terminal voltage of battery $=120-108.5$ $=11.5 \mathrm{~V}$
AP EAMCET-28.04.2017
Current Electricity
152520
A circuit is shown in the figure. The e.m.f. of the battery is
1 $4 \mathrm{~V}$
2 $6 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
C $2 \Omega$ and $1 \Omega$ resistance are in series. $2 \Omega$ resistor through which $1 \mathrm{~A}$ current is flowing, we get- $\mathrm{V}_{\mathrm{CO}}=1 \times 2=2 \mathrm{~V}$ $\therefore \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CO}}}{3}=\frac{2}{3} \mathrm{~A}$ Applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{B}}=1+\mathrm{I}_{\mathrm{C}}=1+\frac{2}{3}=\frac{5}{3} \mathrm{~A}$ $\because \mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{B}} \times 2=\frac{5}{3} \times 2=\frac{10}{3} \mathrm{~V}$ $\therefore \quad \mathrm{V}_{\mathrm{BO}}=\mathrm{V}_{\mathrm{BC}}+\mathrm{V}_{\mathrm{CO}}=\frac{10}{3}+2=\frac{16}{3} \mathrm{~V}$ $\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{V}_{\mathrm{BO}}}{6}=\frac{16 / 3}{6}=\frac{8}{9} \mathrm{~A}$ Again applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}$ $=\frac{5}{3}+\frac{8}{9}=\frac{23}{9} \mathrm{~A}$ $\mathrm{~V}_{\mathrm{AB}}=\mathrm{I}_{\mathrm{A}} \times 2=\frac{23}{9} \times 2=\frac{46}{9} \mathrm{~V}$ Therefore, $\mathrm{V}_{\mathrm{AO}}=\mathrm{V}_{\mathrm{AB}}+\mathrm{V}_{\mathrm{BO}}=\frac{46}{9}+\frac{16}{3}=\frac{94}{9} \mathrm{~V}$ Voltage drop across the battery is the same as the voltage drop between $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$. The emf of the battery $\mathrm{E}=\mathrm{V}_{\mathrm{AO}}=\frac{94}{9}=10.44 \mathrm{~V} \approx 12 \mathrm{~V}$
AMU-2017
Current Electricity
152521
Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is
1 $1: 1$
2 $1: 2$
3 $1: 3$
4 $1: 4$
Explanation:
A Let, the emf of a cell $=\mathrm{E}$ Internal Resistance $=\mathrm{r}$ then in series, $\mathrm{P}_{1}=\left(\frac{2 \mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)^{2} \times 2 \mathrm{r}=2 \mathrm{E}^{2} \mathrm{r}$ In parallel, $\mathrm{P}_{2}=\left(\frac{\mathrm{E}}{\mathrm{r} / 2}\right)^{2} \times \mathrm{r} / 2=2 \mathrm{E}^{2} \mathrm{r}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2 \mathrm{E}^{2} \mathrm{r}}{2 \mathrm{E}^{2} \mathrm{r}}$ $\mathrm{P}_{1}: \mathrm{P}_{2}=1: 1$
AMU-2017
Current Electricity
152523
In the circuit shown below emf of each battery is $5 \mathrm{~V}$; and has an internal resistance of $1.0 \Omega$. The current in the circuit $\left(\mathrm{I}_{0}\right)$ and the reading in an ideal voltmeter $(\mathrm{V})$ are:
1 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=0$
2 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=0$
3 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=1$
4 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=1$
Explanation:
B Given, Internal resistance (r) $=1 \Omega$, And emf $(\mathrm{E})=5 \mathrm{~V}$ Let, $I$ is the current flow in the circuit Net emf of the circuit $=8 \times 5 \mathrm{~V}=40 \mathrm{~V}$ and Net resistance in the circuit $=8 \times 1 \Omega=8 \Omega$ then, current in the circuit $\mathrm{I}=\frac{40}{8}=5 \mathrm{~A}$ And voltmeter reading $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=5-5 \times 1$ $\mathrm{~V}=0$
TS EAMCET(Medical)-2017
Current Electricity
152524
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is connected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is
1 $6 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
D Given, emf of each batteries $(E)=20 \mathrm{~V}$ External Resistance $(\mathrm{R})=4 \Omega$ Voltage drop across $\mathrm{AB}$ and $\mathrm{CD}$ are same i.e. $20 \mathrm{~V}$ Hence, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20}{4}=5 \mathrm{~A}$
152519
A battery of emf $8 \mathrm{~V}$ and internal resistance 0.5 $\Omega$ is being charged by a $120 \mathrm{~V}$ d.c. supply using a series resistor of $15.5 \Omega$. The terminal voltage of $8 \mathrm{~V}$ battery during charging is
1 $11.5 \mathrm{~V}$
2 $112 \mathrm{~V}$
3 $14 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, emf of the storage battery $(\mathrm{E})=8 \mathrm{~V}$ Internal resistance of the battery $(\mathrm{r})=0.5 \Omega$ DC supply voltage $(\mathrm{V})=120 \mathrm{~V}$ Resistance of the resistor $(\mathrm{R})=15.5 \Omega$ $\mathrm{R}$ is connected to the storage battery in series. Effective voltage in the circuit $\left(V^{\prime}\right)=V-E$ $=120-8$ $=112 \mathrm{~V}$ According to Ohm's law, $\mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}+\mathrm{r}}=\frac{112}{15.5+0.5}=\frac{112}{10}=7 \mathrm{~A}$ Voltage across resistance $\mathrm{V} =\mathrm{IR}$ $=7 \times 15.5=108.5 \mathrm{~V}$ DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $\mathrm{R}$ Terminal voltage of battery $=120-108.5$ $=11.5 \mathrm{~V}$
AP EAMCET-28.04.2017
Current Electricity
152520
A circuit is shown in the figure. The e.m.f. of the battery is
1 $4 \mathrm{~V}$
2 $6 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
C $2 \Omega$ and $1 \Omega$ resistance are in series. $2 \Omega$ resistor through which $1 \mathrm{~A}$ current is flowing, we get- $\mathrm{V}_{\mathrm{CO}}=1 \times 2=2 \mathrm{~V}$ $\therefore \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CO}}}{3}=\frac{2}{3} \mathrm{~A}$ Applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{B}}=1+\mathrm{I}_{\mathrm{C}}=1+\frac{2}{3}=\frac{5}{3} \mathrm{~A}$ $\because \mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{B}} \times 2=\frac{5}{3} \times 2=\frac{10}{3} \mathrm{~V}$ $\therefore \quad \mathrm{V}_{\mathrm{BO}}=\mathrm{V}_{\mathrm{BC}}+\mathrm{V}_{\mathrm{CO}}=\frac{10}{3}+2=\frac{16}{3} \mathrm{~V}$ $\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{V}_{\mathrm{BO}}}{6}=\frac{16 / 3}{6}=\frac{8}{9} \mathrm{~A}$ Again applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}$ $=\frac{5}{3}+\frac{8}{9}=\frac{23}{9} \mathrm{~A}$ $\mathrm{~V}_{\mathrm{AB}}=\mathrm{I}_{\mathrm{A}} \times 2=\frac{23}{9} \times 2=\frac{46}{9} \mathrm{~V}$ Therefore, $\mathrm{V}_{\mathrm{AO}}=\mathrm{V}_{\mathrm{AB}}+\mathrm{V}_{\mathrm{BO}}=\frac{46}{9}+\frac{16}{3}=\frac{94}{9} \mathrm{~V}$ Voltage drop across the battery is the same as the voltage drop between $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$. The emf of the battery $\mathrm{E}=\mathrm{V}_{\mathrm{AO}}=\frac{94}{9}=10.44 \mathrm{~V} \approx 12 \mathrm{~V}$
AMU-2017
Current Electricity
152521
Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is
1 $1: 1$
2 $1: 2$
3 $1: 3$
4 $1: 4$
Explanation:
A Let, the emf of a cell $=\mathrm{E}$ Internal Resistance $=\mathrm{r}$ then in series, $\mathrm{P}_{1}=\left(\frac{2 \mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)^{2} \times 2 \mathrm{r}=2 \mathrm{E}^{2} \mathrm{r}$ In parallel, $\mathrm{P}_{2}=\left(\frac{\mathrm{E}}{\mathrm{r} / 2}\right)^{2} \times \mathrm{r} / 2=2 \mathrm{E}^{2} \mathrm{r}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2 \mathrm{E}^{2} \mathrm{r}}{2 \mathrm{E}^{2} \mathrm{r}}$ $\mathrm{P}_{1}: \mathrm{P}_{2}=1: 1$
AMU-2017
Current Electricity
152523
In the circuit shown below emf of each battery is $5 \mathrm{~V}$; and has an internal resistance of $1.0 \Omega$. The current in the circuit $\left(\mathrm{I}_{0}\right)$ and the reading in an ideal voltmeter $(\mathrm{V})$ are:
1 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=0$
2 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=0$
3 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=1$
4 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=1$
Explanation:
B Given, Internal resistance (r) $=1 \Omega$, And emf $(\mathrm{E})=5 \mathrm{~V}$ Let, $I$ is the current flow in the circuit Net emf of the circuit $=8 \times 5 \mathrm{~V}=40 \mathrm{~V}$ and Net resistance in the circuit $=8 \times 1 \Omega=8 \Omega$ then, current in the circuit $\mathrm{I}=\frac{40}{8}=5 \mathrm{~A}$ And voltmeter reading $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=5-5 \times 1$ $\mathrm{~V}=0$
TS EAMCET(Medical)-2017
Current Electricity
152524
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is connected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is
1 $6 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
D Given, emf of each batteries $(E)=20 \mathrm{~V}$ External Resistance $(\mathrm{R})=4 \Omega$ Voltage drop across $\mathrm{AB}$ and $\mathrm{CD}$ are same i.e. $20 \mathrm{~V}$ Hence, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20}{4}=5 \mathrm{~A}$
152519
A battery of emf $8 \mathrm{~V}$ and internal resistance 0.5 $\Omega$ is being charged by a $120 \mathrm{~V}$ d.c. supply using a series resistor of $15.5 \Omega$. The terminal voltage of $8 \mathrm{~V}$ battery during charging is
1 $11.5 \mathrm{~V}$
2 $112 \mathrm{~V}$
3 $14 \mathrm{~V}$
4 $6 \mathrm{~V}$
Explanation:
A Given, emf of the storage battery $(\mathrm{E})=8 \mathrm{~V}$ Internal resistance of the battery $(\mathrm{r})=0.5 \Omega$ DC supply voltage $(\mathrm{V})=120 \mathrm{~V}$ Resistance of the resistor $(\mathrm{R})=15.5 \Omega$ $\mathrm{R}$ is connected to the storage battery in series. Effective voltage in the circuit $\left(V^{\prime}\right)=V-E$ $=120-8$ $=112 \mathrm{~V}$ According to Ohm's law, $\mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}+\mathrm{r}}=\frac{112}{15.5+0.5}=\frac{112}{10}=7 \mathrm{~A}$ Voltage across resistance $\mathrm{V} =\mathrm{IR}$ $=7 \times 15.5=108.5 \mathrm{~V}$ DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $\mathrm{R}$ Terminal voltage of battery $=120-108.5$ $=11.5 \mathrm{~V}$
AP EAMCET-28.04.2017
Current Electricity
152520
A circuit is shown in the figure. The e.m.f. of the battery is
1 $4 \mathrm{~V}$
2 $6 \mathrm{~V}$
3 $12 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
C $2 \Omega$ and $1 \Omega$ resistance are in series. $2 \Omega$ resistor through which $1 \mathrm{~A}$ current is flowing, we get- $\mathrm{V}_{\mathrm{CO}}=1 \times 2=2 \mathrm{~V}$ $\therefore \mathrm{I}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CO}}}{3}=\frac{2}{3} \mathrm{~A}$ Applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{B}}=1+\mathrm{I}_{\mathrm{C}}=1+\frac{2}{3}=\frac{5}{3} \mathrm{~A}$ $\because \mathrm{V}_{\mathrm{BC}}=\mathrm{I}_{\mathrm{B}} \times 2=\frac{5}{3} \times 2=\frac{10}{3} \mathrm{~V}$ $\therefore \quad \mathrm{V}_{\mathrm{BO}}=\mathrm{V}_{\mathrm{BC}}+\mathrm{V}_{\mathrm{CO}}=\frac{10}{3}+2=\frac{16}{3} \mathrm{~V}$ $\mathrm{I}_{\mathrm{D}}=\frac{\mathrm{V}_{\mathrm{BO}}}{6}=\frac{16 / 3}{6}=\frac{8}{9} \mathrm{~A}$ Again applying Kirchhoff's current law, $\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{D}}$ $=\frac{5}{3}+\frac{8}{9}=\frac{23}{9} \mathrm{~A}$ $\mathrm{~V}_{\mathrm{AB}}=\mathrm{I}_{\mathrm{A}} \times 2=\frac{23}{9} \times 2=\frac{46}{9} \mathrm{~V}$ Therefore, $\mathrm{V}_{\mathrm{AO}}=\mathrm{V}_{\mathrm{AB}}+\mathrm{V}_{\mathrm{BO}}=\frac{46}{9}+\frac{16}{3}=\frac{94}{9} \mathrm{~V}$ Voltage drop across the battery is the same as the voltage drop between $\mathrm{V}_{\mathrm{A}}$ and $\mathrm{V}_{\mathrm{B}}$. The emf of the battery $\mathrm{E}=\mathrm{V}_{\mathrm{AO}}=\frac{94}{9}=10.44 \mathrm{~V} \approx 12 \mathrm{~V}$
AMU-2017
Current Electricity
152521
Two identical cells are first connected in series and then in parallel. The ratio of power consumed by them is
1 $1: 1$
2 $1: 2$
3 $1: 3$
4 $1: 4$
Explanation:
A Let, the emf of a cell $=\mathrm{E}$ Internal Resistance $=\mathrm{r}$ then in series, $\mathrm{P}_{1}=\left(\frac{2 \mathrm{E}}{\mathrm{r}+\mathrm{r}}\right)^{2} \times 2 \mathrm{r}=2 \mathrm{E}^{2} \mathrm{r}$ In parallel, $\mathrm{P}_{2}=\left(\frac{\mathrm{E}}{\mathrm{r} / 2}\right)^{2} \times \mathrm{r} / 2=2 \mathrm{E}^{2} \mathrm{r}$ $\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{2 \mathrm{E}^{2} \mathrm{r}}{2 \mathrm{E}^{2} \mathrm{r}}$ $\mathrm{P}_{1}: \mathrm{P}_{2}=1: 1$
AMU-2017
Current Electricity
152523
In the circuit shown below emf of each battery is $5 \mathrm{~V}$; and has an internal resistance of $1.0 \Omega$. The current in the circuit $\left(\mathrm{I}_{0}\right)$ and the reading in an ideal voltmeter $(\mathrm{V})$ are:
1 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=0$
2 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=0$
3 $\mathrm{I}_{0}=1 \mathrm{~A}, \mathrm{~V}=1$
4 $\mathrm{I}_{0}=5 \mathrm{~A}, \mathrm{~V}=1$
Explanation:
B Given, Internal resistance (r) $=1 \Omega$, And emf $(\mathrm{E})=5 \mathrm{~V}$ Let, $I$ is the current flow in the circuit Net emf of the circuit $=8 \times 5 \mathrm{~V}=40 \mathrm{~V}$ and Net resistance in the circuit $=8 \times 1 \Omega=8 \Omega$ then, current in the circuit $\mathrm{I}=\frac{40}{8}=5 \mathrm{~A}$ And voltmeter reading $\mathrm{V}=\mathrm{E}-\mathrm{Ir}$ $\mathrm{V}=5-5 \times 1$ $\mathrm{~V}=0$
TS EAMCET(Medical)-2017
Current Electricity
152524
Each of the six ideal batteries of emf $20 \mathrm{~V}$ is connected to an external resistance of $4 \Omega$ as shown in the figure. The current through the resistance is
1 $6 \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $4 \mathrm{~A}$
4 $5 \mathrm{~A}$
Explanation:
D Given, emf of each batteries $(E)=20 \mathrm{~V}$ External Resistance $(\mathrm{R})=4 \Omega$ Voltage drop across $\mathrm{AB}$ and $\mathrm{CD}$ are same i.e. $20 \mathrm{~V}$ Hence, current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20}{4}=5 \mathrm{~A}$
