NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152489
The emf of a car battery is $12 \mathrm{~V}$ of internal resistance of battery is $0.4 \Omega$ then maximum power drawn battery is W.
1 30
2 360
3 4.8
4 Zero
Explanation:
B Given, emf (E) $=12$ volt Internal resistance $(r)=0.4 \Omega$ We know, Current (i) $=\frac{\text { voltage }}{\text { internal resistance }}=\frac{12}{0.4}=30 \mathrm{~A}$ Power $(\mathrm{P})=\mathrm{VI}$ $=12 \times 30=360 \mathrm{Watt}$
GUJCET 2020
Current Electricity
152491
Six cells, each of emf $5 \mathrm{~V}$ and internal resistance $0.1 \Omega$ are connected as shown in Figure. The reading of the ideal voltmeter $V$ is
1 $30 \mathrm{~V}$
2 $5 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 zero
5 $0.5 \mathrm{~V}$
Explanation:
D The emf of cells connected in reverse polarity cancel to each other. Then the voltmeter reading $=5-5=0$
Kerala CEE 2020
Current Electricity
152493
The current in the $1 \Omega$ resistor shown in the circuit is
1 $\frac{2}{3} \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
D Two $4 \Omega$ resistors are in parallel connection. Their equivalent resistance $\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{R}}=\frac{1}{2}$ $\mathrm{R}=2 \Omega$ $2 \Omega$ and $1 \Omega$ resistor are in series connection. Total Resistance $\mathrm{R}^{\prime}=2+1$ $\mathrm{R}^{\prime}=3 \Omega$ $\therefore$ Current through $1 \Omega$ resistor $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $= \frac{6}{3}$ $I =2 \mathrm{~A}$
VITEEE-2019
Current Electricity
152494
For a battery the electromotive force is $1.5 \mathrm{~V}$. Terminal voltage is $1.25 \mathrm{~V}$. Power supplied to external resistance is $2.5 \mathrm{~W}$. Find the internal resistance of battery
1 $0.125 \Omega$
2 $0.25 \Omega$
3 $0.5 \Omega$
4 $0.1 \Omega$
Explanation:
A Given, $\mathrm{V}=1.25 \mathrm{~V}$ $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{P}_{\mathrm{R}}=2.5 \mathrm{Watt}$ Cell supplies a constant current in the circuit, $\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow 2.5=\frac{(1.25)^{2}}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{(1.25)^{2}}{2.5}$ $\mathrm{R}=0.625 \Omega$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $=\frac{1.25}{0.625}$ $=2 \mathrm{~A}$ Internal resistance of the cell, $r =\left[\frac{\mathrm{E}}{\mathrm{V}}-1\right] \mathrm{R}$ $=\left[\frac{1.5}{1.25}-1\right] \times 0.625$ $=\frac{0.25}{1.25} \times 0.625$ $\mathrm{r} =0.125 \Omega$
152489
The emf of a car battery is $12 \mathrm{~V}$ of internal resistance of battery is $0.4 \Omega$ then maximum power drawn battery is W.
1 30
2 360
3 4.8
4 Zero
Explanation:
B Given, emf (E) $=12$ volt Internal resistance $(r)=0.4 \Omega$ We know, Current (i) $=\frac{\text { voltage }}{\text { internal resistance }}=\frac{12}{0.4}=30 \mathrm{~A}$ Power $(\mathrm{P})=\mathrm{VI}$ $=12 \times 30=360 \mathrm{Watt}$
GUJCET 2020
Current Electricity
152491
Six cells, each of emf $5 \mathrm{~V}$ and internal resistance $0.1 \Omega$ are connected as shown in Figure. The reading of the ideal voltmeter $V$ is
1 $30 \mathrm{~V}$
2 $5 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 zero
5 $0.5 \mathrm{~V}$
Explanation:
D The emf of cells connected in reverse polarity cancel to each other. Then the voltmeter reading $=5-5=0$
Kerala CEE 2020
Current Electricity
152493
The current in the $1 \Omega$ resistor shown in the circuit is
1 $\frac{2}{3} \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
D Two $4 \Omega$ resistors are in parallel connection. Their equivalent resistance $\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{R}}=\frac{1}{2}$ $\mathrm{R}=2 \Omega$ $2 \Omega$ and $1 \Omega$ resistor are in series connection. Total Resistance $\mathrm{R}^{\prime}=2+1$ $\mathrm{R}^{\prime}=3 \Omega$ $\therefore$ Current through $1 \Omega$ resistor $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $= \frac{6}{3}$ $I =2 \mathrm{~A}$
VITEEE-2019
Current Electricity
152494
For a battery the electromotive force is $1.5 \mathrm{~V}$. Terminal voltage is $1.25 \mathrm{~V}$. Power supplied to external resistance is $2.5 \mathrm{~W}$. Find the internal resistance of battery
1 $0.125 \Omega$
2 $0.25 \Omega$
3 $0.5 \Omega$
4 $0.1 \Omega$
Explanation:
A Given, $\mathrm{V}=1.25 \mathrm{~V}$ $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{P}_{\mathrm{R}}=2.5 \mathrm{Watt}$ Cell supplies a constant current in the circuit, $\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow 2.5=\frac{(1.25)^{2}}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{(1.25)^{2}}{2.5}$ $\mathrm{R}=0.625 \Omega$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $=\frac{1.25}{0.625}$ $=2 \mathrm{~A}$ Internal resistance of the cell, $r =\left[\frac{\mathrm{E}}{\mathrm{V}}-1\right] \mathrm{R}$ $=\left[\frac{1.5}{1.25}-1\right] \times 0.625$ $=\frac{0.25}{1.25} \times 0.625$ $\mathrm{r} =0.125 \Omega$
152489
The emf of a car battery is $12 \mathrm{~V}$ of internal resistance of battery is $0.4 \Omega$ then maximum power drawn battery is W.
1 30
2 360
3 4.8
4 Zero
Explanation:
B Given, emf (E) $=12$ volt Internal resistance $(r)=0.4 \Omega$ We know, Current (i) $=\frac{\text { voltage }}{\text { internal resistance }}=\frac{12}{0.4}=30 \mathrm{~A}$ Power $(\mathrm{P})=\mathrm{VI}$ $=12 \times 30=360 \mathrm{Watt}$
GUJCET 2020
Current Electricity
152491
Six cells, each of emf $5 \mathrm{~V}$ and internal resistance $0.1 \Omega$ are connected as shown in Figure. The reading of the ideal voltmeter $V$ is
1 $30 \mathrm{~V}$
2 $5 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 zero
5 $0.5 \mathrm{~V}$
Explanation:
D The emf of cells connected in reverse polarity cancel to each other. Then the voltmeter reading $=5-5=0$
Kerala CEE 2020
Current Electricity
152493
The current in the $1 \Omega$ resistor shown in the circuit is
1 $\frac{2}{3} \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
D Two $4 \Omega$ resistors are in parallel connection. Their equivalent resistance $\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{R}}=\frac{1}{2}$ $\mathrm{R}=2 \Omega$ $2 \Omega$ and $1 \Omega$ resistor are in series connection. Total Resistance $\mathrm{R}^{\prime}=2+1$ $\mathrm{R}^{\prime}=3 \Omega$ $\therefore$ Current through $1 \Omega$ resistor $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $= \frac{6}{3}$ $I =2 \mathrm{~A}$
VITEEE-2019
Current Electricity
152494
For a battery the electromotive force is $1.5 \mathrm{~V}$. Terminal voltage is $1.25 \mathrm{~V}$. Power supplied to external resistance is $2.5 \mathrm{~W}$. Find the internal resistance of battery
1 $0.125 \Omega$
2 $0.25 \Omega$
3 $0.5 \Omega$
4 $0.1 \Omega$
Explanation:
A Given, $\mathrm{V}=1.25 \mathrm{~V}$ $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{P}_{\mathrm{R}}=2.5 \mathrm{Watt}$ Cell supplies a constant current in the circuit, $\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow 2.5=\frac{(1.25)^{2}}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{(1.25)^{2}}{2.5}$ $\mathrm{R}=0.625 \Omega$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $=\frac{1.25}{0.625}$ $=2 \mathrm{~A}$ Internal resistance of the cell, $r =\left[\frac{\mathrm{E}}{\mathrm{V}}-1\right] \mathrm{R}$ $=\left[\frac{1.5}{1.25}-1\right] \times 0.625$ $=\frac{0.25}{1.25} \times 0.625$ $\mathrm{r} =0.125 \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
152489
The emf of a car battery is $12 \mathrm{~V}$ of internal resistance of battery is $0.4 \Omega$ then maximum power drawn battery is W.
1 30
2 360
3 4.8
4 Zero
Explanation:
B Given, emf (E) $=12$ volt Internal resistance $(r)=0.4 \Omega$ We know, Current (i) $=\frac{\text { voltage }}{\text { internal resistance }}=\frac{12}{0.4}=30 \mathrm{~A}$ Power $(\mathrm{P})=\mathrm{VI}$ $=12 \times 30=360 \mathrm{Watt}$
GUJCET 2020
Current Electricity
152491
Six cells, each of emf $5 \mathrm{~V}$ and internal resistance $0.1 \Omega$ are connected as shown in Figure. The reading of the ideal voltmeter $V$ is
1 $30 \mathrm{~V}$
2 $5 \mathrm{~V}$
3 $15 \mathrm{~V}$
4 zero
5 $0.5 \mathrm{~V}$
Explanation:
D The emf of cells connected in reverse polarity cancel to each other. Then the voltmeter reading $=5-5=0$
Kerala CEE 2020
Current Electricity
152493
The current in the $1 \Omega$ resistor shown in the circuit is
1 $\frac{2}{3} \mathrm{~A}$
2 $3 \mathrm{~A}$
3 $6 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
D Two $4 \Omega$ resistors are in parallel connection. Their equivalent resistance $\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{R}}=\frac{1}{2}$ $\mathrm{R}=2 \Omega$ $2 \Omega$ and $1 \Omega$ resistor are in series connection. Total Resistance $\mathrm{R}^{\prime}=2+1$ $\mathrm{R}^{\prime}=3 \Omega$ $\therefore$ Current through $1 \Omega$ resistor $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}$ $= \frac{6}{3}$ $I =2 \mathrm{~A}$
VITEEE-2019
Current Electricity
152494
For a battery the electromotive force is $1.5 \mathrm{~V}$. Terminal voltage is $1.25 \mathrm{~V}$. Power supplied to external resistance is $2.5 \mathrm{~W}$. Find the internal resistance of battery
1 $0.125 \Omega$
2 $0.25 \Omega$
3 $0.5 \Omega$
4 $0.1 \Omega$
Explanation:
A Given, $\mathrm{V}=1.25 \mathrm{~V}$ $\mathrm{E}=1.5 \mathrm{~V}$ $\mathrm{P}_{\mathrm{R}}=2.5 \mathrm{Watt}$ Cell supplies a constant current in the circuit, $\mathrm{P}_{\mathrm{R}}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow 2.5=\frac{(1.25)^{2}}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{(1.25)^{2}}{2.5}$ $\mathrm{R}=0.625 \Omega$ Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $=\frac{1.25}{0.625}$ $=2 \mathrm{~A}$ Internal resistance of the cell, $r =\left[\frac{\mathrm{E}}{\mathrm{V}}-1\right] \mathrm{R}$ $=\left[\frac{1.5}{1.25}-1\right] \times 0.625$ $=\frac{0.25}{1.25} \times 0.625$ $\mathrm{r} =0.125 \Omega$
