(B) Let $V_{P1} = 0\,V$. Then the potential at $P_2$ is $V$. Using nodal analysis at node $P_2$:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 2}{1} = 0$

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(B) Let $V_{P1} = 0\,V$. Then the potential at $P_2$ is $V$. Using nodal analysis at node $P_2$:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 2}{1} = 0$Multiply by $10$:$5(V - 5) + V + 10(V - 2) = 0$$5V - 25 + V + 10V - 20 = 0$$16V = 45$$V = \frac{45}{16} = 2.8125\,V$The current through the $10\,\Omega$ resistor is $I = \frac{V}{R} = \frac{2.8125}{10} = 0.28125\,A \approx 0.28\,A$.Given the options,the closest value is $0.27\,A$. Since $V_{P2} > V_{P1}$,the current flows from $P_2$ to $P_1$.

Class 12 Physics Current Electricity Circuit Solving for current and Voltage

Medium

$A$ $5\ V$ battery with internal resistance $2\,\Omega$ and a $2\,V$ battery with internal resistance $1\,\Omega$ are connected to a $10\,\Omega$ resistor as shown in the figure. Find the current through the $10\,\Omega$ resistor.

AIEEE 2008 All AIEEE

A
$0.27\,A$,$P_1$ to $P_2$
B
$0.27\,A$,$P_2$ to $P_1$
C
$0.03\,A$,$P_1$ to $P_2$
D
$0.03\,A$,$P_2$ to $P_1$

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Circuit Solving for current and Voltage

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$A$ $5\ V$ battery with internal resistance $2\,\Omega$ and a $2\,V$ battery with internal resistance $1\,\Omega$ are connected to a $10\,\Omega$ resistor as shown in the figure. Find the current through the $10\,\Omega$ resistor.

Similar Questions

In the circuit shown in the figure,the ammeter reads a current of ............. $A$.

In the circuit shown,the power developed in the $6\,\Omega$ resistor is $6\,W$. The power developed in the $4\,\Omega$ resistor is .............. $W$.

For a cell, the graph between the potential difference $(V)$ across the terminals of the cell and the current $(I)$ drawn from the cell is shown in the figure. The $e.m.f.$ and the internal resistance of the cell are

Twelve wires each having resistance $2 \Omega$ are joined to form a cube. $A$ battery of $6 \text{ V}$ emf is connected across points $a$ and $c$. The voltage difference between $e$ and $f$ is . . . . . . $\text{V}$.

In the circuit shown in the figure,if the potential at point $A$ is taken to be zero,the potential at point $B$ is ................ $V$.

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