Question

In the circuit shown in figure, calculate the following:

(a) Potential difference between points a and b when switch S is open.

(b) Current through S in the circuit when S is closed.

Answer 1

(a) -12 V, (b) 3 A (flowing from b to a)

Answer 2

(a) When the switch S is open, points 'a' and 'b' are disconnected. The circuit consists of two independent voltage dividers. For point 'a': The potential at point 'a' ($V_a$) is calculated using the voltage divider rule: $V_a = 36 \, V \times \frac{3\,\Omega}{6\,\Omega + 3\,\Omega} = 12 \, V$ For point 'b': The potential at point 'b' ($V_b$) is calculated using the voltage divider rule: $V_b = 36 \, V \times \frac{6\,\Omega}{3\,\Omega + 6\,\Omega} = 24 \, V$ The potential difference between points 'a' and 'b' is $V_{ab} = V_a - V_b = 12 \, V - 24 \, V = -12 \, V$.

(b) When the switch S is closed, points 'a' and 'b' are connected, meaning $V_a = V_b$. Let this common potential be $V_x$. Applying Kirchhoff's Current Law (KCL) at the node $V_x$: $\frac{36 - V_x}{6} + \frac{36 - V_x}{3} = \frac{V_x}{3} + \frac{V_x}{6}$ Multiplying by 6: $(36 - V_x) + 2(36 - V_x) = 2V_x + V_x$ $108 - 3V_x = 3V_x$ $108 = 6V_x$ $V_x = 18 \, V$ Now, find the current through the switch S. Let $I_{a \to b}$ be the current flowing from 'a' to 'b' through the switch. Applying KCL at node 'a': $\frac{36 - V_a}{6} = \frac{V_a}{3} + I_{a \to b}$ $3 \, A = 6 \, A + I_{a \to b}$ $I_{a \to b} = 3 \, A - 6 \, A = -3 \, A$ The negative sign indicates that the current flows from 'b' to 'a' with a magnitude of 3A.