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Current Electricity

Question

Wired Faculty App

A 5 amp fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is

0.2 Ω

5 Ω

0.4 Ω

0.04 Ω

Solution

D.

0.04 Ω

Electric power = volltage difference × current

P = i V

= i × iR

⇒ Power P = i² R

⇒ R = $\frac{P}{i^{2}}$

Given:- P = 1W and i = 5A

∴ R = $\frac{1}{{(5)}^{2}}$

R = 0.04 Ω

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