Question

For the series parallel arrangement shown in below figure find a) the supply current b) the current flowing through each resistor and c) the potential difference across each resistor
Pls refer attachment for diagram

Answer 1

Answer:

here's your answer mate

Answer 2

Answer:

We can see that R₂ and R₃ are connected in parallel networks.

So the resultant resistance, Rp = $\frac{R3 * R2}{R3 + R2}$ = $\frac{6 * 2}{6 + 2}$ = 12/8 = 1.5 ohm

and  R₄ is connected with Rp as well as R₁ in series.

So equivalent total resistance : R =  R₄ + R₁ + Rp = 2.5 + 4 + 1.5 = 8ohm

a) For supply current,

we know that V = IR (OHM'S LAW)

or, V = 200 V (given)

or, I = V / R = 200 / 8 = 25A

b) Current flowing through each resistor,

along R₁ and R₄ the current will stay the same since it is connect in a series combination

So, current through R₄ and R₁ will be 25A

current across the parallel combination will be 25A. So the current gets equally divided into two equal amount.

c) Potential difference across the section from R₁ and R₄ will  be

Rv = R₄ - R₁ = 4 - 2.5 = 1.5 ohm

So by OHM'SM LAW:

V = IR = 1.5 x 25 = 37.5V

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