Current Electricity: Numerical Problems Set (7)

Question

Three resistors $R_1 = 2~\Omega$, $R_2 = 3~\Omega$ and $R_3 = 6~\Omega$ are connected. $R_2$ and $R_3$ are in parallel, and that combination is in series with $R_1$. The network is connected to a battery of EMF $12$ V with internal resistance $r = 1~\Omega$. Find (a) the current drawn from the battery, (b) the potential drop across the parallel combination and (c) the power dissipated in $R_3$.

Solution — Step by Step

Why This Works

Two CBSE-board moves drive every mixed circuit: (1) collapse parallel pairs into a single equivalent, then (2) treat the simplified network as one big series loop where Ohm’s law gives the main-line current. After that, voltages and individual-branch currents fall out one by one.

The power formula $P = V^2/R$ is the cleanest choice when you know the voltage across the resistor. If you only know the current through it, use $P = I^2 R$. Both give the same answer — pick whichever avoids extra arithmetic.

Alternative Method

Find the current through $R_3$ directly. The voltage $V_p = 4.8$ V drives $R_3$, so $I_3 = 4.8/6 = 0.8$ A. Then $P = I^2 R = (0.8)^2 \times 6 = 3.84$ W. Same answer.

Common Mistake