Current Electricity and Electrical InstrumentMedium
Question
A wire of length ' $l$ ' and resistance $100\Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
Options
A.$55\Omega$
B.$60\Omega$
C.$26\Omega$
D.$52\Omega$
Solution
Sol. [Key Concept: Resistance of a wire cut into equal parts (R ∝ length)]
Divided into 10 parts
$${R = \frac{\rho l}{A} }{R^{'} = \frac{\rho l}{10A} = \frac{R}{10} }$$$R_{S} = 5 \times \frac{R}{10}\ $ [series]
$$R_{S} = 50 $$$R_{P} = \frac{R}{50}\ $ [parallel]
$$\begin{matrix} R_{eq} & \ = R_{S} + R_{P} \\ & \ = 52\Omega \end{matrix}$$
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