Current Electricity Ques 95
95. Two rods are joined end to end, as shown. Both have a cross-sectional area of $0.01 cm^{2}$. Each is 1 meter long. One rod is of copper with a resistivity of $1.7 \times 10^{-6}$ ohm-centimeter, the other is of iron with a resistivity of $10^{-5} ohm-$ centimeter.
How much voltage is required to produce a current of 1 ampere in the rods?
(a) $0.117 V$
(b) $0.00145 V$
(c) $0.0145 V$
(d) $1.7 \times 10^{-6} V$
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Answer:
Correct Answer: 95.(a)
Solution:
- (a) Copper rod and iron rod are joined in series.
$\therefore R=R _{Cu}+R _{Fe}=(\rho_1+\rho_2) \frac{\ell}{A} \quad(\because R=\rho \frac{\ell}{A})$
From ohm’s law $V=R I$
$=(1.7 \times 10^{-6} \times 10^{-2}+10^{-5} \times 10^{-2})$ $\div 0.01 \times 10^{-4}$ volt
$=0.117 volt(\because I=1 A)$
BETA
