Question

Calculate the current drawn from the battery by the network of four

resistors shown in the figure.l​

Answer 1

When the circuit is balanced , current across the galvanometer ( not given here )  will be zero .

From top ,

• 2 Ω , 2 Ω are connected in series ,

➠ $\sf R_{1eq}=2\Omega +2\Omega$

➠ $\sf \orange{R_{1eq}=4\Omega}\ \; \bigstar$

From below too ,

• 2 Ω , 2 Ω are connected in series ,

➤ $\sf R_{2eq}=2\Omega +2\Omega$

➤ $\sf \green{R_{2eq}=4\Omega}\ \; \bigstar$

Now , $\sf R_{1eq}\ ,R_{2eq}$ are connected in parallel ,

➙ $\sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_{1eq}}+\dfrac{1}{R_{2eq}}$

➙ $\sf \dfrac{1}{R_{eq}}=\dfrac{1}{4}+\dfrac{1}{4}$

➙  $\sf \dfrac{1}{R_{eq}}=\dfrac{2}{4}$

➙  $\sf R_{eq}=\dfrac{4}{2}$

➙  $\sf R_{eq}=2\ \Omega\ \; \pink{\bigstar}$

Apply Ohm's law ,

⭆ $\sf V=IR_{eq}$

⭆ 4 = I (2)

⭆ I = 4 / 2

⭆ I = 2 A  $\blue{\bigstar}$

So , Current drawn from the battery is 2 A .