Question

Please tell answer with explanation.

Answer 1

Answer:

i can't understand the question sorry

Answer 2

$\huge{\bold{\underline{\underline{Solution}}}}$

Consider the following parallel circuit shown below: Let $\bold{I_{1}, \: I_{2} \: and \: I_{3}}$ be the current flow through the resistor $\bold{R_{1}, \: R_{2} \: and \: R_{3}}$ connected in parallel. Using Ohm’s law, current through each resistor is:

$\bold{I = I_{1} + I_{2} + I_{3}}$

$\bold{V = V/R_{1} + V/R_{2} + V/R_{3}}$

$\bold{V/R_{p} = V(1/R_{1} + 1/R_{2} + 1/R_{3}})$

$\bold{1/R_{p} = V(1/R_{1} + 1/R_{2} + 1/R_{3}})$

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• (a) R5 and R4 with Parallel combination of R2 and R3 are in series

• (b) R2 and R3 are in parallel.

• (c) R2 and R3 in parallel gives Rp = 1 Ω Rp, R5 and R4 are in series. So, Req = 5 Ω R1 is not to be taken as it is shorted. Current flowing, l = V/R = 5/5 = 1A.

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Hope it helps uh<3