152628
A battery of emf 2.0 volts and internal resistance $0.10 \Omega$ is being charged with a current of 5.0 A. The potential difference between the terminals of the battery is
1 $2.0 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 zero
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=2 \mathrm{~V}$ Internal Resistance $(\mathrm{r})=0.1 \Omega$ Current $(\mathrm{I})=5 \mathrm{~A}$ We know potential difference when cell being charged $\mathrm{V}=\mathrm{E}+\mathrm{Ir}$ $\mathrm{V}=2+0.1 \times 5$ $\mathrm{~V}=2.5 \mathrm{~V}$
MP PET-2012
Current Electricity
152629
Two cells having the internal resistance $0.2 \Omega$ and $0.4 \Omega$ are connected in parallel. The voltage across the battery terminal is 1.5 Volt. The e.m.f. of first cell is 1.2 Volt. The e.m.f. of the second cell is
1 2.7 Volt
2 2.1 Volt
3 3 Volt
4 4.2 Volt
Explanation:
B Given $r_{1}=0.2 \Omega, r_{2}=0.4 \Omega$ and $\mathrm{V}=1.5 \mathrm{~V}$ and $\mathrm{E}_{1}=1.2 \mathrm{~V}$ From ohm's law $\mathrm{V}=\mathrm{IR}$ Then, $\quad R_{\mathrm{eq}}=\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$ and current $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ $\mathrm{V}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+}\right.$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1}+}{\mathrm{r}_{1}} \mathrm{r}_{1}+\mathrm{r}_{2}\right.$ $\mathrm{V}=\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{1.5=\frac{1.2 \times 0.4+0.2 \times}{0.2+0.4}}$ $0.9=0.48+0.2 \mathrm{E}_{2}$ $0.2 \mathrm{E}_{2}=0.9-0.48$ $0.2 \mathrm{E}_{2}=0.42$ $\mathrm{E}_{2}=\frac{0.42}{0.2}$ $\mathrm{E}_{2}=2.1 \mathrm{~V}$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right) \quad\left(\because \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $1.5=\frac{1.2 \times 0.4+0.2 \times \mathrm{E}_{2}}{0.2+0.4}$
MP PET-2008
Current Electricity
152630
The internal resistance of a primary cell is $4 \Omega$. It generates a current of $0.2 \mathrm{~A}$ in an external resistance of $21 \Omega$. The rate at which chemical energy to consumed in providing current is
1 $1 \mathrm{~J} / \mathrm{s}$
2 $5 \mathrm{~J} / \mathrm{s}$
3 $0.42 \mathrm{~J} / \mathrm{s}$
4 $0.8 \mathrm{~J} / \mathrm{s}$
Explanation:
A Given Internal resistance of primary cell $(\mathrm{r})=4 \Omega$ Current $(\mathrm{I})=0.2 \mathrm{~A}$ External resistance $(\mathrm{R})=21 \Omega$ Equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=4+21=25 \Omega$ We know that, Rate of generation electrical energy $\left(P=I^{2} R\right)=$ Rate of consumption of chemical energy Consumption of chemical energy $=(0.2)^{2} \times 25$ $=0.4 \times 25$ $=1.00 \mathrm{~J} / \mathrm{s}$
Current Electricity
152631
A battery of emf $10 \mathrm{~V}$ and internal resistance of $0.5 \mathrm{ohm}$ is connected across a variable resistance $R$. The maximum value of $R$ is given by
1 $0.5 \Omega$
2 $1.00 \Omega$
3 $2.0 \Omega$
4 $0.25 \Omega$
Explanation:
A Given, $\operatorname{emf}(\mathrm{E})=10 \mathrm{~V}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ External resistance $=\mathrm{R}$ We know, $I=\frac{E}{R+r}=\frac{10}{R+0.5}=\frac{20}{2 R+1}$ And $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\left(\frac{20}{2 \mathrm{R}+1}\right)^{2} \mathrm{R}$ $=\frac{400}{(2 \mathrm{R}+1)^{2}} \mathrm{R}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\frac{\mathrm{dP}}{\mathrm{dR}}=400\left[\frac{(2 \mathrm{R}+1)^{2}-4 \mathrm{R}(2 \mathrm{R}+1)}{(2 \mathrm{R}+1)^{4}}\right]=0$ $(2 \mathrm{R}+1)^{2}=4 \mathrm{R}(2 \mathrm{R}+1)$ $2 \mathrm{R}+1=4 \mathrm{R}$ $2 \mathrm{R}=1$ $\mathrm{R}=0.5 \Omega$
152628
A battery of emf 2.0 volts and internal resistance $0.10 \Omega$ is being charged with a current of 5.0 A. The potential difference between the terminals of the battery is
1 $2.0 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 zero
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=2 \mathrm{~V}$ Internal Resistance $(\mathrm{r})=0.1 \Omega$ Current $(\mathrm{I})=5 \mathrm{~A}$ We know potential difference when cell being charged $\mathrm{V}=\mathrm{E}+\mathrm{Ir}$ $\mathrm{V}=2+0.1 \times 5$ $\mathrm{~V}=2.5 \mathrm{~V}$
MP PET-2012
Current Electricity
152629
Two cells having the internal resistance $0.2 \Omega$ and $0.4 \Omega$ are connected in parallel. The voltage across the battery terminal is 1.5 Volt. The e.m.f. of first cell is 1.2 Volt. The e.m.f. of the second cell is
1 2.7 Volt
2 2.1 Volt
3 3 Volt
4 4.2 Volt
Explanation:
B Given $r_{1}=0.2 \Omega, r_{2}=0.4 \Omega$ and $\mathrm{V}=1.5 \mathrm{~V}$ and $\mathrm{E}_{1}=1.2 \mathrm{~V}$ From ohm's law $\mathrm{V}=\mathrm{IR}$ Then, $\quad R_{\mathrm{eq}}=\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$ and current $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ $\mathrm{V}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+}\right.$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1}+}{\mathrm{r}_{1}} \mathrm{r}_{1}+\mathrm{r}_{2}\right.$ $\mathrm{V}=\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{1.5=\frac{1.2 \times 0.4+0.2 \times}{0.2+0.4}}$ $0.9=0.48+0.2 \mathrm{E}_{2}$ $0.2 \mathrm{E}_{2}=0.9-0.48$ $0.2 \mathrm{E}_{2}=0.42$ $\mathrm{E}_{2}=\frac{0.42}{0.2}$ $\mathrm{E}_{2}=2.1 \mathrm{~V}$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right) \quad\left(\because \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $1.5=\frac{1.2 \times 0.4+0.2 \times \mathrm{E}_{2}}{0.2+0.4}$
MP PET-2008
Current Electricity
152630
The internal resistance of a primary cell is $4 \Omega$. It generates a current of $0.2 \mathrm{~A}$ in an external resistance of $21 \Omega$. The rate at which chemical energy to consumed in providing current is
1 $1 \mathrm{~J} / \mathrm{s}$
2 $5 \mathrm{~J} / \mathrm{s}$
3 $0.42 \mathrm{~J} / \mathrm{s}$
4 $0.8 \mathrm{~J} / \mathrm{s}$
Explanation:
A Given Internal resistance of primary cell $(\mathrm{r})=4 \Omega$ Current $(\mathrm{I})=0.2 \mathrm{~A}$ External resistance $(\mathrm{R})=21 \Omega$ Equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=4+21=25 \Omega$ We know that, Rate of generation electrical energy $\left(P=I^{2} R\right)=$ Rate of consumption of chemical energy Consumption of chemical energy $=(0.2)^{2} \times 25$ $=0.4 \times 25$ $=1.00 \mathrm{~J} / \mathrm{s}$
Current Electricity
152631
A battery of emf $10 \mathrm{~V}$ and internal resistance of $0.5 \mathrm{ohm}$ is connected across a variable resistance $R$. The maximum value of $R$ is given by
1 $0.5 \Omega$
2 $1.00 \Omega$
3 $2.0 \Omega$
4 $0.25 \Omega$
Explanation:
A Given, $\operatorname{emf}(\mathrm{E})=10 \mathrm{~V}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ External resistance $=\mathrm{R}$ We know, $I=\frac{E}{R+r}=\frac{10}{R+0.5}=\frac{20}{2 R+1}$ And $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\left(\frac{20}{2 \mathrm{R}+1}\right)^{2} \mathrm{R}$ $=\frac{400}{(2 \mathrm{R}+1)^{2}} \mathrm{R}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\frac{\mathrm{dP}}{\mathrm{dR}}=400\left[\frac{(2 \mathrm{R}+1)^{2}-4 \mathrm{R}(2 \mathrm{R}+1)}{(2 \mathrm{R}+1)^{4}}\right]=0$ $(2 \mathrm{R}+1)^{2}=4 \mathrm{R}(2 \mathrm{R}+1)$ $2 \mathrm{R}+1=4 \mathrm{R}$ $2 \mathrm{R}=1$ $\mathrm{R}=0.5 \Omega$
152628
A battery of emf 2.0 volts and internal resistance $0.10 \Omega$ is being charged with a current of 5.0 A. The potential difference between the terminals of the battery is
1 $2.0 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 zero
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=2 \mathrm{~V}$ Internal Resistance $(\mathrm{r})=0.1 \Omega$ Current $(\mathrm{I})=5 \mathrm{~A}$ We know potential difference when cell being charged $\mathrm{V}=\mathrm{E}+\mathrm{Ir}$ $\mathrm{V}=2+0.1 \times 5$ $\mathrm{~V}=2.5 \mathrm{~V}$
MP PET-2012
Current Electricity
152629
Two cells having the internal resistance $0.2 \Omega$ and $0.4 \Omega$ are connected in parallel. The voltage across the battery terminal is 1.5 Volt. The e.m.f. of first cell is 1.2 Volt. The e.m.f. of the second cell is
1 2.7 Volt
2 2.1 Volt
3 3 Volt
4 4.2 Volt
Explanation:
B Given $r_{1}=0.2 \Omega, r_{2}=0.4 \Omega$ and $\mathrm{V}=1.5 \mathrm{~V}$ and $\mathrm{E}_{1}=1.2 \mathrm{~V}$ From ohm's law $\mathrm{V}=\mathrm{IR}$ Then, $\quad R_{\mathrm{eq}}=\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$ and current $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ $\mathrm{V}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+}\right.$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1}+}{\mathrm{r}_{1}} \mathrm{r}_{1}+\mathrm{r}_{2}\right.$ $\mathrm{V}=\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{1.5=\frac{1.2 \times 0.4+0.2 \times}{0.2+0.4}}$ $0.9=0.48+0.2 \mathrm{E}_{2}$ $0.2 \mathrm{E}_{2}=0.9-0.48$ $0.2 \mathrm{E}_{2}=0.42$ $\mathrm{E}_{2}=\frac{0.42}{0.2}$ $\mathrm{E}_{2}=2.1 \mathrm{~V}$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right) \quad\left(\because \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $1.5=\frac{1.2 \times 0.4+0.2 \times \mathrm{E}_{2}}{0.2+0.4}$
MP PET-2008
Current Electricity
152630
The internal resistance of a primary cell is $4 \Omega$. It generates a current of $0.2 \mathrm{~A}$ in an external resistance of $21 \Omega$. The rate at which chemical energy to consumed in providing current is
1 $1 \mathrm{~J} / \mathrm{s}$
2 $5 \mathrm{~J} / \mathrm{s}$
3 $0.42 \mathrm{~J} / \mathrm{s}$
4 $0.8 \mathrm{~J} / \mathrm{s}$
Explanation:
A Given Internal resistance of primary cell $(\mathrm{r})=4 \Omega$ Current $(\mathrm{I})=0.2 \mathrm{~A}$ External resistance $(\mathrm{R})=21 \Omega$ Equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=4+21=25 \Omega$ We know that, Rate of generation electrical energy $\left(P=I^{2} R\right)=$ Rate of consumption of chemical energy Consumption of chemical energy $=(0.2)^{2} \times 25$ $=0.4 \times 25$ $=1.00 \mathrm{~J} / \mathrm{s}$
Current Electricity
152631
A battery of emf $10 \mathrm{~V}$ and internal resistance of $0.5 \mathrm{ohm}$ is connected across a variable resistance $R$. The maximum value of $R$ is given by
1 $0.5 \Omega$
2 $1.00 \Omega$
3 $2.0 \Omega$
4 $0.25 \Omega$
Explanation:
A Given, $\operatorname{emf}(\mathrm{E})=10 \mathrm{~V}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ External resistance $=\mathrm{R}$ We know, $I=\frac{E}{R+r}=\frac{10}{R+0.5}=\frac{20}{2 R+1}$ And $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\left(\frac{20}{2 \mathrm{R}+1}\right)^{2} \mathrm{R}$ $=\frac{400}{(2 \mathrm{R}+1)^{2}} \mathrm{R}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\frac{\mathrm{dP}}{\mathrm{dR}}=400\left[\frac{(2 \mathrm{R}+1)^{2}-4 \mathrm{R}(2 \mathrm{R}+1)}{(2 \mathrm{R}+1)^{4}}\right]=0$ $(2 \mathrm{R}+1)^{2}=4 \mathrm{R}(2 \mathrm{R}+1)$ $2 \mathrm{R}+1=4 \mathrm{R}$ $2 \mathrm{R}=1$ $\mathrm{R}=0.5 \Omega$
152628
A battery of emf 2.0 volts and internal resistance $0.10 \Omega$ is being charged with a current of 5.0 A. The potential difference between the terminals of the battery is
1 $2.0 \mathrm{~V}$
2 $2.5 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 zero
Explanation:
B Given, $\operatorname{Emf}(\mathrm{E})=2 \mathrm{~V}$ Internal Resistance $(\mathrm{r})=0.1 \Omega$ Current $(\mathrm{I})=5 \mathrm{~A}$ We know potential difference when cell being charged $\mathrm{V}=\mathrm{E}+\mathrm{Ir}$ $\mathrm{V}=2+0.1 \times 5$ $\mathrm{~V}=2.5 \mathrm{~V}$
MP PET-2012
Current Electricity
152629
Two cells having the internal resistance $0.2 \Omega$ and $0.4 \Omega$ are connected in parallel. The voltage across the battery terminal is 1.5 Volt. The e.m.f. of first cell is 1.2 Volt. The e.m.f. of the second cell is
1 2.7 Volt
2 2.1 Volt
3 3 Volt
4 4.2 Volt
Explanation:
B Given $r_{1}=0.2 \Omega, r_{2}=0.4 \Omega$ and $\mathrm{V}=1.5 \mathrm{~V}$ and $\mathrm{E}_{1}=1.2 \mathrm{~V}$ From ohm's law $\mathrm{V}=\mathrm{IR}$ Then, $\quad R_{\mathrm{eq}}=\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$ and current $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ $\mathrm{V}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+}\right.$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1}+}{\mathrm{r}_{1}} \mathrm{r}_{1}+\mathrm{r}_{2}\right.$ $\mathrm{V}=\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{1.5=\frac{1.2 \times 0.4+0.2 \times}{0.2+0.4}}$ $0.9=0.48+0.2 \mathrm{E}_{2}$ $0.2 \mathrm{E}_{2}=0.9-0.48$ $0.2 \mathrm{E}_{2}=0.42$ $\mathrm{E}_{2}=\frac{0.42}{0.2}$ $\mathrm{E}_{2}=2.1 \mathrm{~V}$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right) \quad\left(\because \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}}\right)$ $\mathrm{V}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1} \mathrm{r}_{2}}\right)\left(\frac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ $1.5=\frac{1.2 \times 0.4+0.2 \times \mathrm{E}_{2}}{0.2+0.4}$
MP PET-2008
Current Electricity
152630
The internal resistance of a primary cell is $4 \Omega$. It generates a current of $0.2 \mathrm{~A}$ in an external resistance of $21 \Omega$. The rate at which chemical energy to consumed in providing current is
1 $1 \mathrm{~J} / \mathrm{s}$
2 $5 \mathrm{~J} / \mathrm{s}$
3 $0.42 \mathrm{~J} / \mathrm{s}$
4 $0.8 \mathrm{~J} / \mathrm{s}$
Explanation:
A Given Internal resistance of primary cell $(\mathrm{r})=4 \Omega$ Current $(\mathrm{I})=0.2 \mathrm{~A}$ External resistance $(\mathrm{R})=21 \Omega$ Equivalent resistance $\left(\mathrm{R}_{\mathrm{eq}}\right)=4+21=25 \Omega$ We know that, Rate of generation electrical energy $\left(P=I^{2} R\right)=$ Rate of consumption of chemical energy Consumption of chemical energy $=(0.2)^{2} \times 25$ $=0.4 \times 25$ $=1.00 \mathrm{~J} / \mathrm{s}$
Current Electricity
152631
A battery of emf $10 \mathrm{~V}$ and internal resistance of $0.5 \mathrm{ohm}$ is connected across a variable resistance $R$. The maximum value of $R$ is given by
1 $0.5 \Omega$
2 $1.00 \Omega$
3 $2.0 \Omega$
4 $0.25 \Omega$
Explanation:
A Given, $\operatorname{emf}(\mathrm{E})=10 \mathrm{~V}$ Internal resistance $(\mathrm{r})=0.5 \Omega$ External resistance $=\mathrm{R}$ We know, $I=\frac{E}{R+r}=\frac{10}{R+0.5}=\frac{20}{2 R+1}$ And $\text { Power }(\mathrm{P})=\mathrm{I}^{2} \mathrm{R}$ $=\left(\frac{20}{2 \mathrm{R}+1}\right)^{2} \mathrm{R}$ $=\frac{400}{(2 \mathrm{R}+1)^{2}} \mathrm{R}$ For maximum power, $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ $\frac{\mathrm{dP}}{\mathrm{dR}}=400\left[\frac{(2 \mathrm{R}+1)^{2}-4 \mathrm{R}(2 \mathrm{R}+1)}{(2 \mathrm{R}+1)^{4}}\right]=0$ $(2 \mathrm{R}+1)^{2}=4 \mathrm{R}(2 \mathrm{R}+1)$ $2 \mathrm{R}+1=4 \mathrm{R}$ $2 \mathrm{R}=1$ $\mathrm{R}=0.5 \Omega$
