152505
Find out the current $I_{2}$ as shown in the diagram.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.1 \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
A Applying Kirchhoff's voltage law for loop (1) $10-\mathrm{I}_{2} \times 20-10 \times \mathrm{i}=0$ $1-2 \mathrm{I}_{2}-\mathrm{i}=0$ Applying Kirchhoff's voltage law for loop (2) $10-\mathrm{I}_{2} \times 20-\left(\mathrm{I}_{2}-\mathrm{i}\right) \times 10=0$ $1-2 \mathrm{I}_{2}-\mathrm{I}_{2}+\mathrm{i}=0$ $1-3 \mathrm{I}_{2}+\mathrm{i}=0$ Adding equation (i) and equation (ii), we have $2-5 \mathrm{I}_{2}=0$ $5 \mathrm{I}_{2}=2$ $\mathrm{I}_{2}=\frac{2}{5}$ $\mathrm{I}_{2}=0.4 \mathrm{~A}$
AIIMS-27.05.2018(M)
Current Electricity
152506
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
1 $0.27 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
2 $0.03 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
3 $0.03 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
4 $0.27 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
Explanation:
C Let potential at $P_{1}$ is $0 \mathrm{~V}$ and potential at $\mathrm{P}_{2}$ is $\mathrm{V}_{0}$ Applying KCL at $\mathrm{P}_{2}$ $\frac{\mathrm{V}_{0}-5}{2}+\frac{\mathrm{V}_{0}-0}{10}+\frac{\mathrm{V}_{0}-(-2)}{1}=0$ $\mathrm{~V}_{0}=\frac{5}{16}$ So, current through $10 \Omega$ resistor is, $\mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}=\frac{\frac{5}{16}}{10}=\frac{1}{32} \mathrm{~A}$ $\mathrm{I}=0.03125 \approx 0.03 \mathrm{~A}\left(\text { From } \mathrm{P}_{2} \text { to } \mathrm{P}_{1}\right)$
AIIMS-26.05.2018(E)
Current Electricity
152508
Assertion: Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emf. Reason: The equivalent internal resistance is greater than either of the two internal resistances.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
D The equivalent emf of the two batteries in parallel $\mathrm{E}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ Here, $\mathrm{E}$ may be; $\mathrm{E}_{1} \leq \mathrm{E} \leq \mathrm{E}_{2}$ And equivalent internal resistance of two batteries connected in parallel, $r=\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$ This value of $r$ is smaller than either of $r_{1}$ and $r_{2}$.
AIIMS-27.05.2018(E)]
Current Electricity
152509
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. The current in the circuit is $0.5 \mathrm{~A}$. The terminal voltage of the battery when the circuit is closed is
1 $10 \mathrm{~V}$
2 $0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $8.5 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{i}=0.5 \mathrm{~A}$ Resistance of the resistor $i=\frac{E}{R+r}$ $0.5=\frac{10}{R+3}$ $R=17 \Omega$ According to ohm's law $\mathrm{V} =\mathrm{iR}$ $=0.5 \times 17=8.5 \text { Volt }$
AIIMS-27.05.2018(M)]
Current Electricity
152510
A $10 \mathrm{~V}$ battery with internal resistance $1 \Omega$ and a $15 \mathrm{~V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (sec figure). The reading in the voltmeter will be close to : $\text { (1) }$
1 $12.5 \mathrm{~V}$
2 $24.5 \mathrm{~V}$
3 $13.1 \mathrm{~V}$
4 $11.9 \mathrm{~V}$
Explanation:
C Using Kirchhoff's voltage law in the given circuit $15-10-1 \times \mathrm{i}-0.6 \times \mathrm{i}=0$ $5=1.6 \mathrm{i}$ $\mathrm{i}=3.125 \mathrm{~A}$ By Kirchhoff's Voltage law, $\Delta \mathrm{V}=15-0.6 \mathrm{i}$ $=15-0.6 \times 3.125$ $=13.125$ Voltage
152505
Find out the current $I_{2}$ as shown in the diagram.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.1 \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
A Applying Kirchhoff's voltage law for loop (1) $10-\mathrm{I}_{2} \times 20-10 \times \mathrm{i}=0$ $1-2 \mathrm{I}_{2}-\mathrm{i}=0$ Applying Kirchhoff's voltage law for loop (2) $10-\mathrm{I}_{2} \times 20-\left(\mathrm{I}_{2}-\mathrm{i}\right) \times 10=0$ $1-2 \mathrm{I}_{2}-\mathrm{I}_{2}+\mathrm{i}=0$ $1-3 \mathrm{I}_{2}+\mathrm{i}=0$ Adding equation (i) and equation (ii), we have $2-5 \mathrm{I}_{2}=0$ $5 \mathrm{I}_{2}=2$ $\mathrm{I}_{2}=\frac{2}{5}$ $\mathrm{I}_{2}=0.4 \mathrm{~A}$
AIIMS-27.05.2018(M)
Current Electricity
152506
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
1 $0.27 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
2 $0.03 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
3 $0.03 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
4 $0.27 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
Explanation:
C Let potential at $P_{1}$ is $0 \mathrm{~V}$ and potential at $\mathrm{P}_{2}$ is $\mathrm{V}_{0}$ Applying KCL at $\mathrm{P}_{2}$ $\frac{\mathrm{V}_{0}-5}{2}+\frac{\mathrm{V}_{0}-0}{10}+\frac{\mathrm{V}_{0}-(-2)}{1}=0$ $\mathrm{~V}_{0}=\frac{5}{16}$ So, current through $10 \Omega$ resistor is, $\mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}=\frac{\frac{5}{16}}{10}=\frac{1}{32} \mathrm{~A}$ $\mathrm{I}=0.03125 \approx 0.03 \mathrm{~A}\left(\text { From } \mathrm{P}_{2} \text { to } \mathrm{P}_{1}\right)$
AIIMS-26.05.2018(E)
Current Electricity
152508
Assertion: Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emf. Reason: The equivalent internal resistance is greater than either of the two internal resistances.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
D The equivalent emf of the two batteries in parallel $\mathrm{E}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ Here, $\mathrm{E}$ may be; $\mathrm{E}_{1} \leq \mathrm{E} \leq \mathrm{E}_{2}$ And equivalent internal resistance of two batteries connected in parallel, $r=\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$ This value of $r$ is smaller than either of $r_{1}$ and $r_{2}$.
AIIMS-27.05.2018(E)]
Current Electricity
152509
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. The current in the circuit is $0.5 \mathrm{~A}$. The terminal voltage of the battery when the circuit is closed is
1 $10 \mathrm{~V}$
2 $0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $8.5 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{i}=0.5 \mathrm{~A}$ Resistance of the resistor $i=\frac{E}{R+r}$ $0.5=\frac{10}{R+3}$ $R=17 \Omega$ According to ohm's law $\mathrm{V} =\mathrm{iR}$ $=0.5 \times 17=8.5 \text { Volt }$
AIIMS-27.05.2018(M)]
Current Electricity
152510
A $10 \mathrm{~V}$ battery with internal resistance $1 \Omega$ and a $15 \mathrm{~V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (sec figure). The reading in the voltmeter will be close to : $\text { (1) }$
1 $12.5 \mathrm{~V}$
2 $24.5 \mathrm{~V}$
3 $13.1 \mathrm{~V}$
4 $11.9 \mathrm{~V}$
Explanation:
C Using Kirchhoff's voltage law in the given circuit $15-10-1 \times \mathrm{i}-0.6 \times \mathrm{i}=0$ $5=1.6 \mathrm{i}$ $\mathrm{i}=3.125 \mathrm{~A}$ By Kirchhoff's Voltage law, $\Delta \mathrm{V}=15-0.6 \mathrm{i}$ $=15-0.6 \times 3.125$ $=13.125$ Voltage
152505
Find out the current $I_{2}$ as shown in the diagram.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.1 \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
A Applying Kirchhoff's voltage law for loop (1) $10-\mathrm{I}_{2} \times 20-10 \times \mathrm{i}=0$ $1-2 \mathrm{I}_{2}-\mathrm{i}=0$ Applying Kirchhoff's voltage law for loop (2) $10-\mathrm{I}_{2} \times 20-\left(\mathrm{I}_{2}-\mathrm{i}\right) \times 10=0$ $1-2 \mathrm{I}_{2}-\mathrm{I}_{2}+\mathrm{i}=0$ $1-3 \mathrm{I}_{2}+\mathrm{i}=0$ Adding equation (i) and equation (ii), we have $2-5 \mathrm{I}_{2}=0$ $5 \mathrm{I}_{2}=2$ $\mathrm{I}_{2}=\frac{2}{5}$ $\mathrm{I}_{2}=0.4 \mathrm{~A}$
AIIMS-27.05.2018(M)
Current Electricity
152506
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
1 $0.27 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
2 $0.03 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
3 $0.03 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
4 $0.27 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
Explanation:
C Let potential at $P_{1}$ is $0 \mathrm{~V}$ and potential at $\mathrm{P}_{2}$ is $\mathrm{V}_{0}$ Applying KCL at $\mathrm{P}_{2}$ $\frac{\mathrm{V}_{0}-5}{2}+\frac{\mathrm{V}_{0}-0}{10}+\frac{\mathrm{V}_{0}-(-2)}{1}=0$ $\mathrm{~V}_{0}=\frac{5}{16}$ So, current through $10 \Omega$ resistor is, $\mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}=\frac{\frac{5}{16}}{10}=\frac{1}{32} \mathrm{~A}$ $\mathrm{I}=0.03125 \approx 0.03 \mathrm{~A}\left(\text { From } \mathrm{P}_{2} \text { to } \mathrm{P}_{1}\right)$
AIIMS-26.05.2018(E)
Current Electricity
152508
Assertion: Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emf. Reason: The equivalent internal resistance is greater than either of the two internal resistances.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
D The equivalent emf of the two batteries in parallel $\mathrm{E}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ Here, $\mathrm{E}$ may be; $\mathrm{E}_{1} \leq \mathrm{E} \leq \mathrm{E}_{2}$ And equivalent internal resistance of two batteries connected in parallel, $r=\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$ This value of $r$ is smaller than either of $r_{1}$ and $r_{2}$.
AIIMS-27.05.2018(E)]
Current Electricity
152509
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. The current in the circuit is $0.5 \mathrm{~A}$. The terminal voltage of the battery when the circuit is closed is
1 $10 \mathrm{~V}$
2 $0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $8.5 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{i}=0.5 \mathrm{~A}$ Resistance of the resistor $i=\frac{E}{R+r}$ $0.5=\frac{10}{R+3}$ $R=17 \Omega$ According to ohm's law $\mathrm{V} =\mathrm{iR}$ $=0.5 \times 17=8.5 \text { Volt }$
AIIMS-27.05.2018(M)]
Current Electricity
152510
A $10 \mathrm{~V}$ battery with internal resistance $1 \Omega$ and a $15 \mathrm{~V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (sec figure). The reading in the voltmeter will be close to : $\text { (1) }$
1 $12.5 \mathrm{~V}$
2 $24.5 \mathrm{~V}$
3 $13.1 \mathrm{~V}$
4 $11.9 \mathrm{~V}$
Explanation:
C Using Kirchhoff's voltage law in the given circuit $15-10-1 \times \mathrm{i}-0.6 \times \mathrm{i}=0$ $5=1.6 \mathrm{i}$ $\mathrm{i}=3.125 \mathrm{~A}$ By Kirchhoff's Voltage law, $\Delta \mathrm{V}=15-0.6 \mathrm{i}$ $=15-0.6 \times 3.125$ $=13.125$ Voltage
152505
Find out the current $I_{2}$ as shown in the diagram.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.1 \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
A Applying Kirchhoff's voltage law for loop (1) $10-\mathrm{I}_{2} \times 20-10 \times \mathrm{i}=0$ $1-2 \mathrm{I}_{2}-\mathrm{i}=0$ Applying Kirchhoff's voltage law for loop (2) $10-\mathrm{I}_{2} \times 20-\left(\mathrm{I}_{2}-\mathrm{i}\right) \times 10=0$ $1-2 \mathrm{I}_{2}-\mathrm{I}_{2}+\mathrm{i}=0$ $1-3 \mathrm{I}_{2}+\mathrm{i}=0$ Adding equation (i) and equation (ii), we have $2-5 \mathrm{I}_{2}=0$ $5 \mathrm{I}_{2}=2$ $\mathrm{I}_{2}=\frac{2}{5}$ $\mathrm{I}_{2}=0.4 \mathrm{~A}$
AIIMS-27.05.2018(M)
Current Electricity
152506
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
1 $0.27 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
2 $0.03 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
3 $0.03 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
4 $0.27 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
Explanation:
C Let potential at $P_{1}$ is $0 \mathrm{~V}$ and potential at $\mathrm{P}_{2}$ is $\mathrm{V}_{0}$ Applying KCL at $\mathrm{P}_{2}$ $\frac{\mathrm{V}_{0}-5}{2}+\frac{\mathrm{V}_{0}-0}{10}+\frac{\mathrm{V}_{0}-(-2)}{1}=0$ $\mathrm{~V}_{0}=\frac{5}{16}$ So, current through $10 \Omega$ resistor is, $\mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}=\frac{\frac{5}{16}}{10}=\frac{1}{32} \mathrm{~A}$ $\mathrm{I}=0.03125 \approx 0.03 \mathrm{~A}\left(\text { From } \mathrm{P}_{2} \text { to } \mathrm{P}_{1}\right)$
AIIMS-26.05.2018(E)
Current Electricity
152508
Assertion: Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emf. Reason: The equivalent internal resistance is greater than either of the two internal resistances.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
D The equivalent emf of the two batteries in parallel $\mathrm{E}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ Here, $\mathrm{E}$ may be; $\mathrm{E}_{1} \leq \mathrm{E} \leq \mathrm{E}_{2}$ And equivalent internal resistance of two batteries connected in parallel, $r=\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$ This value of $r$ is smaller than either of $r_{1}$ and $r_{2}$.
AIIMS-27.05.2018(E)]
Current Electricity
152509
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. The current in the circuit is $0.5 \mathrm{~A}$. The terminal voltage of the battery when the circuit is closed is
1 $10 \mathrm{~V}$
2 $0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $8.5 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{i}=0.5 \mathrm{~A}$ Resistance of the resistor $i=\frac{E}{R+r}$ $0.5=\frac{10}{R+3}$ $R=17 \Omega$ According to ohm's law $\mathrm{V} =\mathrm{iR}$ $=0.5 \times 17=8.5 \text { Volt }$
AIIMS-27.05.2018(M)]
Current Electricity
152510
A $10 \mathrm{~V}$ battery with internal resistance $1 \Omega$ and a $15 \mathrm{~V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (sec figure). The reading in the voltmeter will be close to : $\text { (1) }$
1 $12.5 \mathrm{~V}$
2 $24.5 \mathrm{~V}$
3 $13.1 \mathrm{~V}$
4 $11.9 \mathrm{~V}$
Explanation:
C Using Kirchhoff's voltage law in the given circuit $15-10-1 \times \mathrm{i}-0.6 \times \mathrm{i}=0$ $5=1.6 \mathrm{i}$ $\mathrm{i}=3.125 \mathrm{~A}$ By Kirchhoff's Voltage law, $\Delta \mathrm{V}=15-0.6 \mathrm{i}$ $=15-0.6 \times 3.125$ $=13.125$ Voltage
152505
Find out the current $I_{2}$ as shown in the diagram.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.1 \mathrm{~A}$
4 $1 \mathrm{~A}$
Explanation:
A Applying Kirchhoff's voltage law for loop (1) $10-\mathrm{I}_{2} \times 20-10 \times \mathrm{i}=0$ $1-2 \mathrm{I}_{2}-\mathrm{i}=0$ Applying Kirchhoff's voltage law for loop (2) $10-\mathrm{I}_{2} \times 20-\left(\mathrm{I}_{2}-\mathrm{i}\right) \times 10=0$ $1-2 \mathrm{I}_{2}-\mathrm{I}_{2}+\mathrm{i}=0$ $1-3 \mathrm{I}_{2}+\mathrm{i}=0$ Adding equation (i) and equation (ii), we have $2-5 \mathrm{I}_{2}=0$ $5 \mathrm{I}_{2}=2$ $\mathrm{I}_{2}=\frac{2}{5}$ $\mathrm{I}_{2}=0.4 \mathrm{~A}$
AIIMS-27.05.2018(M)
Current Electricity
152506
A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
1 $0.27 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
2 $0.03 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
3 $0.03 \mathrm{AP}_{2}$ to $\mathrm{P}_{1}$
4 $0.27 \mathrm{AP}_{1}$ to $\mathrm{P}_{2}$
Explanation:
C Let potential at $P_{1}$ is $0 \mathrm{~V}$ and potential at $\mathrm{P}_{2}$ is $\mathrm{V}_{0}$ Applying KCL at $\mathrm{P}_{2}$ $\frac{\mathrm{V}_{0}-5}{2}+\frac{\mathrm{V}_{0}-0}{10}+\frac{\mathrm{V}_{0}-(-2)}{1}=0$ $\mathrm{~V}_{0}=\frac{5}{16}$ So, current through $10 \Omega$ resistor is, $\mathrm{V}=\mathrm{IR}$ $\mathrm{I}=\frac{\Delta \mathrm{V}}{\mathrm{R}}=\frac{\frac{5}{16}}{10}=\frac{1}{32} \mathrm{~A}$ $\mathrm{I}=0.03125 \approx 0.03 \mathrm{~A}\left(\text { From } \mathrm{P}_{2} \text { to } \mathrm{P}_{1}\right)$
AIIMS-26.05.2018(E)
Current Electricity
152508
Assertion: Two non-ideal batteries are connected in parallel. The equivalent emf is smaller than either of the two emf. Reason: The equivalent internal resistance is greater than either of the two internal resistances.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
D The equivalent emf of the two batteries in parallel $\mathrm{E}=\left(\frac{\mathrm{E}_{1} \mathrm{r}_{2}+\mathrm{E}_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$ Here, $\mathrm{E}$ may be; $\mathrm{E}_{1} \leq \mathrm{E} \leq \mathrm{E}_{2}$ And equivalent internal resistance of two batteries connected in parallel, $r=\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$ This value of $r$ is smaller than either of $r_{1}$ and $r_{2}$.
AIIMS-27.05.2018(E)]
Current Electricity
152509
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. The current in the circuit is $0.5 \mathrm{~A}$. The terminal voltage of the battery when the circuit is closed is
1 $10 \mathrm{~V}$
2 $0 \mathrm{~V}$
3 $1.5 \mathrm{~V}$
4 $8.5 \mathrm{~V}$
Explanation:
D Given $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{i}=0.5 \mathrm{~A}$ Resistance of the resistor $i=\frac{E}{R+r}$ $0.5=\frac{10}{R+3}$ $R=17 \Omega$ According to ohm's law $\mathrm{V} =\mathrm{iR}$ $=0.5 \times 17=8.5 \text { Volt }$
AIIMS-27.05.2018(M)]
Current Electricity
152510
A $10 \mathrm{~V}$ battery with internal resistance $1 \Omega$ and a $15 \mathrm{~V}$ battery with internal resistance $0.6 \Omega$ are connected in parallel to a voltmeter (sec figure). The reading in the voltmeter will be close to : $\text { (1) }$
1 $12.5 \mathrm{~V}$
2 $24.5 \mathrm{~V}$
3 $13.1 \mathrm{~V}$
4 $11.9 \mathrm{~V}$
Explanation:
C Using Kirchhoff's voltage law in the given circuit $15-10-1 \times \mathrm{i}-0.6 \times \mathrm{i}=0$ $5=1.6 \mathrm{i}$ $\mathrm{i}=3.125 \mathrm{~A}$ By Kirchhoff's Voltage law, $\Delta \mathrm{V}=15-0.6 \mathrm{i}$ $=15-0.6 \times 3.125$ $=13.125$ Voltage
