Three resistors are joined together as shown in the figure. The resistors are connected to an ammeter and to a cell of emf 9.0 V

Calculate

(a) the effective resistance of the circuit.

(b) the current drawn from the cell.

In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 Ω and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R$p} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{2}{6} \\[0.5em] Rp = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]

Hence, the effective resistance of the circuit = 3 Ω

(b) current drawn = ?

R = 3 Ω

V = 9.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

9 = I x 3
⇒ I = $\dfrac{9}{3}$ = 3 A

Hence, the current drawn = 3 A