Question

of V
(a) With the help of a diagram, derive the formula for the resultant resistance of three resistors connected in
series.
(b) For the circuit shown in the diagram given below :
52
ws
1022
302
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6 V​

Answer 1

Answer:

hope this helps u

Explanation:

(a) Fig. shows three resistances R1, R2 and R3 Connected in series with a battery of let the p.d. across R1, R2 and R3 is V1, V2 and V3 respectively.

S.t. V = V1 + V3 + V2 -------------(1)

Let the equivalent resistance be R and current flowing through whole circuit is 1.

By ohm’s law,

V/1

V = I × R -----------(2)

Applying ohm’s law to both R1, R2 and R3,

V1 = I × R1 --------(3)

V2 = I × R2 ---------(4)

V3 = I × R3 ----------(5)

From eqs. (1), (2), (3), (4) and (5), we get

I × R = I × R1 + I × R2 + I × R3

I × R = I × (R1 + R2 + R3)

R = R1 + R2 + R3

(b) Let 5 Ω = R1, 10 Ω = R2, 30 Ω = R3

(i) Current through R1 = I1 = V/R1 = 6/5 = 1.2A

Current through R2 = I2 = V/R2 = 6/10 = 0.6 A

Current through R3 = I3 = V/R3 = 6/30 = 0. 2 A

(ii) Total current in the circuit = 1.2 + 0.6 + 0.2 = 2A

(iii) Effective resistance R is given as

1/R = 1/R1 + 1/R2 + 1/R3

= 1/5 + 1/10 + 1/30

= (6 + 3 + 1)/30 = 10/30

R = 30/10 = 3 Ω