Study the following circuit :

On the basis of this circuit, answer the following questions :

(a)

Find the value of total resistance between the points A and B.

(b)

Find the resistance between the points B and C.

(c)

(i)OR(c)

Calculate the current drawn from the battery, when the key is closed.

(a) \(\displaystyle \mathrm{R}_{\mathrm{S}}=4 \Omega+6 \Omega+16 \Omega=26 \Omega\)

(b) \(\displaystyle \frac{1}{R_{P}}=\frac{1}{8 \Omega}+\frac{1}{8 \Omega}=\frac{1}{4} \Omega\)

(c) (i) Total resistance \(\displaystyle =26 \Omega+4 \Omega=30 \Omega\) Potential difference \(\displaystyle =\mathrm{V}=6 \mathrm{~V}\) \[\begin{aligned} & \text { Current } \mathrm{I}=\frac{V}{R} \\ & \frac{6}{30}=\frac{1}{5} \mathrm{~A} \quad \text { or } 0.2 \mathrm{~A} . \end{aligned} \](c)(ii) \(\displaystyle 16 \Omega\) Justification: According to Ohm's law when same current flows, the potential difference across a higher resistance is always higher./ Potential difference across \(\displaystyle 16 \Omega=\mathrm{V}=\mathrm{IR} \quad=0.2 \times 16=3.2 \mathrm{~V}\) Potential difference across \(\displaystyle 8 \Omega=\mathrm{V}=\mathrm{IR}_{(\text {total })}=0.2 \times 4=0.8 \mathrm{~V}\)