Question

Two cells each of 5 Volt are connected in series across a 8 ohm resistor and three parallel resistors of 4ohm , 6 ohm , 12 ohm.Draw a circuit diagram for the above arrangement.Calculate the current drawn by the cell and calculate the current through each resistor.

Answer 1

Net voltage of two batteries, connected in series is $U = 10 V$. Let $R_{par}$ denote the net resistance of three resistors, connected in parallel. According to the law of total resistance in parallel circuit, $\frac{1}{R_{par}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ , from where $R_{par} = 2 \Omega$ . Total resistance of all 4 resistors ($R_0$ connected in series with $R_{par}$) is: $R = R_0 + R_{par} = 10 \Omega$.

Hence, according to Ohm's law, the current, going from the batteries is $I = \frac{U}{R} = \frac{10 V}{10 \Omega} = 1 A$.

The same current is going through $R_0$ and $R_{par}$. The voltage on $R_{par}$ is $U_{par} = I R_{par} = 2 V$. Hence,

1. $I_0 = 1 A$
2. $I_1 = \frac{U_{par}}{R_1} = 0.5 A$
3. $I_2 = \frac{U_{par}}{R_2} = \frac{1}{3} A$
4. $I_3 = \frac{U_{par}}{R_3} = \frac{1}{6} A$