152525
A cell of negligible internal resistance and emf $2.5 \mathrm{~V}$ is connected to a series combination of 1 $\Omega, 1 \Omega$ and $3 \Omega$. The potential difference across 3 $\Omega$ resistance is
1 $7.5 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.5 \mathrm{~V}$
Explanation:
C Equivalent resistance, $\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$ $=1+1+3$ $=5 \Omega$ By Ohm's law, $\therefore \quad \mathrm{V} =\mathrm{IR}_{\mathrm{eq}}$ $\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{2.5}{5}$ $\mathrm{I} =0.5 \mathrm{~A}$ Potential drop across $3 \Omega$ resistor $=\mathrm{IR}_{3}=0.5 \times 3=1.5 \mathrm{~V}$
J and K-CET-2017
Current Electricity
152526
Resistances $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. If a $1.5 \mathrm{~V}$ cell of negligible internal resistance is connected across the $3 \Omega$ resistor, the current flowing through this resistor will be
1 $0.25 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $1.0 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Here, Resistance $1 \Omega$ and $2 \Omega$ are connected in series and these two resistance are parallel with $3 \Omega$ resistance. Hence, voltage drop across the $3 \Omega$ and $3 \Omega(1 \Omega$ and $2 \Omega)$ resistance are same. $\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ $\mathrm{R}=1.5 \Omega$ $\text { Electricity }$ Current Electricity Ohm's law $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{1.5}{1.5}=1 \mathrm{~A}$ Hence, the current through the $3 \Omega$ resistor $=\frac{1}{2}=0.5 \mathrm{~A}$
VITEEE-2017
Current Electricity
152527
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$. is
1 $0.2 \Omega$
2 $0.5 \Omega$
3 $0.8 \Omega$
4 $1.0 \Omega$
Explanation:
B Given, $\mathrm{E}=2.1 \mathrm{~V}$ $\mathrm{I}=0.2 \mathrm{~A} \text { and }$ $\mathrm{R}=10 \Omega$ $\mathrm{r}=?$ We know that, $I=\frac{E}{R+r}$ $0.2=\frac{2.1}{10+r}$ $2+0.2 r=2.1$ $0.2 r=0.1$ $r=\frac{1}{2}=0.5 \Omega$ $r=0.5 \Omega$
NEET-2013
Current Electricity
152529
The cells has an emf of $2 \mathrm{~V}$ and the internal resistance of this cell is $0.1 \Omega$, it is connected to resistance of $3.9 \Omega$, the voltage across the cell will be
152525
A cell of negligible internal resistance and emf $2.5 \mathrm{~V}$ is connected to a series combination of 1 $\Omega, 1 \Omega$ and $3 \Omega$. The potential difference across 3 $\Omega$ resistance is
1 $7.5 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.5 \mathrm{~V}$
Explanation:
C Equivalent resistance, $\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$ $=1+1+3$ $=5 \Omega$ By Ohm's law, $\therefore \quad \mathrm{V} =\mathrm{IR}_{\mathrm{eq}}$ $\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{2.5}{5}$ $\mathrm{I} =0.5 \mathrm{~A}$ Potential drop across $3 \Omega$ resistor $=\mathrm{IR}_{3}=0.5 \times 3=1.5 \mathrm{~V}$
J and K-CET-2017
Current Electricity
152526
Resistances $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. If a $1.5 \mathrm{~V}$ cell of negligible internal resistance is connected across the $3 \Omega$ resistor, the current flowing through this resistor will be
1 $0.25 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $1.0 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Here, Resistance $1 \Omega$ and $2 \Omega$ are connected in series and these two resistance are parallel with $3 \Omega$ resistance. Hence, voltage drop across the $3 \Omega$ and $3 \Omega(1 \Omega$ and $2 \Omega)$ resistance are same. $\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ $\mathrm{R}=1.5 \Omega$ $\text { Electricity }$ Current Electricity Ohm's law $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{1.5}{1.5}=1 \mathrm{~A}$ Hence, the current through the $3 \Omega$ resistor $=\frac{1}{2}=0.5 \mathrm{~A}$
VITEEE-2017
Current Electricity
152527
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$. is
1 $0.2 \Omega$
2 $0.5 \Omega$
3 $0.8 \Omega$
4 $1.0 \Omega$
Explanation:
B Given, $\mathrm{E}=2.1 \mathrm{~V}$ $\mathrm{I}=0.2 \mathrm{~A} \text { and }$ $\mathrm{R}=10 \Omega$ $\mathrm{r}=?$ We know that, $I=\frac{E}{R+r}$ $0.2=\frac{2.1}{10+r}$ $2+0.2 r=2.1$ $0.2 r=0.1$ $r=\frac{1}{2}=0.5 \Omega$ $r=0.5 \Omega$
NEET-2013
Current Electricity
152529
The cells has an emf of $2 \mathrm{~V}$ and the internal resistance of this cell is $0.1 \Omega$, it is connected to resistance of $3.9 \Omega$, the voltage across the cell will be
152525
A cell of negligible internal resistance and emf $2.5 \mathrm{~V}$ is connected to a series combination of 1 $\Omega, 1 \Omega$ and $3 \Omega$. The potential difference across 3 $\Omega$ resistance is
1 $7.5 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.5 \mathrm{~V}$
Explanation:
C Equivalent resistance, $\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$ $=1+1+3$ $=5 \Omega$ By Ohm's law, $\therefore \quad \mathrm{V} =\mathrm{IR}_{\mathrm{eq}}$ $\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{2.5}{5}$ $\mathrm{I} =0.5 \mathrm{~A}$ Potential drop across $3 \Omega$ resistor $=\mathrm{IR}_{3}=0.5 \times 3=1.5 \mathrm{~V}$
J and K-CET-2017
Current Electricity
152526
Resistances $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. If a $1.5 \mathrm{~V}$ cell of negligible internal resistance is connected across the $3 \Omega$ resistor, the current flowing through this resistor will be
1 $0.25 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $1.0 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Here, Resistance $1 \Omega$ and $2 \Omega$ are connected in series and these two resistance are parallel with $3 \Omega$ resistance. Hence, voltage drop across the $3 \Omega$ and $3 \Omega(1 \Omega$ and $2 \Omega)$ resistance are same. $\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ $\mathrm{R}=1.5 \Omega$ $\text { Electricity }$ Current Electricity Ohm's law $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{1.5}{1.5}=1 \mathrm{~A}$ Hence, the current through the $3 \Omega$ resistor $=\frac{1}{2}=0.5 \mathrm{~A}$
VITEEE-2017
Current Electricity
152527
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$. is
1 $0.2 \Omega$
2 $0.5 \Omega$
3 $0.8 \Omega$
4 $1.0 \Omega$
Explanation:
B Given, $\mathrm{E}=2.1 \mathrm{~V}$ $\mathrm{I}=0.2 \mathrm{~A} \text { and }$ $\mathrm{R}=10 \Omega$ $\mathrm{r}=?$ We know that, $I=\frac{E}{R+r}$ $0.2=\frac{2.1}{10+r}$ $2+0.2 r=2.1$ $0.2 r=0.1$ $r=\frac{1}{2}=0.5 \Omega$ $r=0.5 \Omega$
NEET-2013
Current Electricity
152529
The cells has an emf of $2 \mathrm{~V}$ and the internal resistance of this cell is $0.1 \Omega$, it is connected to resistance of $3.9 \Omega$, the voltage across the cell will be
NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
152525
A cell of negligible internal resistance and emf $2.5 \mathrm{~V}$ is connected to a series combination of 1 $\Omega, 1 \Omega$ and $3 \Omega$. The potential difference across 3 $\Omega$ resistance is
1 $7.5 \mathrm{~V}$
2 $2 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $2.5 \mathrm{~V}$
Explanation:
C Equivalent resistance, $\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$ $=1+1+3$ $=5 \Omega$ By Ohm's law, $\therefore \quad \mathrm{V} =\mathrm{IR}_{\mathrm{eq}}$ $\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{2.5}{5}$ $\mathrm{I} =0.5 \mathrm{~A}$ Potential drop across $3 \Omega$ resistor $=\mathrm{IR}_{3}=0.5 \times 3=1.5 \mathrm{~V}$
J and K-CET-2017
Current Electricity
152526
Resistances $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. If a $1.5 \mathrm{~V}$ cell of negligible internal resistance is connected across the $3 \Omega$ resistor, the current flowing through this resistor will be
1 $0.25 \mathrm{~A}$
2 $0.5 \mathrm{~A}$
3 $1.0 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
B Here, Resistance $1 \Omega$ and $2 \Omega$ are connected in series and these two resistance are parallel with $3 \Omega$ resistance. Hence, voltage drop across the $3 \Omega$ and $3 \Omega(1 \Omega$ and $2 \Omega)$ resistance are same. $\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ $\mathrm{R}=1.5 \Omega$ $\text { Electricity }$ Current Electricity Ohm's law $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{1.5}{1.5}=1 \mathrm{~A}$ Hence, the current through the $3 \Omega$ resistor $=\frac{1}{2}=0.5 \mathrm{~A}$
VITEEE-2017
Current Electricity
152527
The internal resistance of a $2.1 \mathrm{~V}$ cell which gives a current of $0.2 \mathrm{~A}$ through a resistance of $10 \Omega$. is
1 $0.2 \Omega$
2 $0.5 \Omega$
3 $0.8 \Omega$
4 $1.0 \Omega$
Explanation:
B Given, $\mathrm{E}=2.1 \mathrm{~V}$ $\mathrm{I}=0.2 \mathrm{~A} \text { and }$ $\mathrm{R}=10 \Omega$ $\mathrm{r}=?$ We know that, $I=\frac{E}{R+r}$ $0.2=\frac{2.1}{10+r}$ $2+0.2 r=2.1$ $0.2 r=0.1$ $r=\frac{1}{2}=0.5 \Omega$ $r=0.5 \Omega$
NEET-2013
Current Electricity
152529
The cells has an emf of $2 \mathrm{~V}$ and the internal resistance of this cell is $0.1 \Omega$, it is connected to resistance of $3.9 \Omega$, the voltage across the cell will be
