Q.Two cells, each of emf 5 V, are connected in series with an $8\ \Omega$ resistor and three parallel resistors of $4\ \Omega$, $6\ \Omega$ and $12\ \Omega$. Draw a circuit diagram for the above arrangement. Calculate

Step 1. Two 5 V cells in series give a combined emf of $5+5=10$ V (internal resistance not given, taken as negligible).

Step 2. The three parallel resistors combine as $\dfrac{1}{R_P}=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}+\dfrac{2}{12}+\dfrac{1}{12}=\dfrac{6}{12}=\dfrac12$, so $R_P=2\ \Omega$.

Step 3. This parallel combination is in series with the $8\ \Omega$ resistor, giving total circuit resistance $R_{tot}=8+2=10\ \Omega$.

Step 4 (i). The current drawn from the cells is $I=\dfrac{\varepsilon_{total}}{R_{tot}}=\dfrac{10}{10}=1$ A.

Step 5. The voltage across the parallel combination is $V_P=IR_P=1\times2=2$ V (this is the common voltage across all three parallel resistors). …

Combine the three parallel resistors into R_P, add the series 8 Ω, find the total current, then find the comm …

• Forgetting to add the two 5 V cells' emfs (10 V total) before computing the current. …