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Question 17

When the switch $$S$$, in the circuit shown, is closed, then the value of current $$i$$ will be:

Solution

Let the potential at node $$C$$ be $$V_C$$.

Applying Kirchhoff's Current Law at node $$C$$: $$i_1 + i_2 = i$$

$$\frac{20 - V_C}{2} + \frac{10 - V_C}{4} = \frac{V_C - 0}{2}$$

$$2(20 - V_C) + (10 - V_C) = 2V_C$$

$$40 - 2V_C + 10 - V_C = 2V_C \implies 50 = 5V_C \implies V_C = 10\text{ V}$$

$$i = \frac{V_C}{2} = \frac{10}{2} = 5\text{ A}$$

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