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In R1 = 10 Omega, R2 = 40 Omega and R3 =...

In `R_1 = 10 Omega, R_2 = 40 Omega and R_3 = 30 Omega, R_4 = 20 Omega, R_5 = 60 Omega` and a `12 V` battery is connected to the arrangement, calculate.
(a) the total resistance and
(b) the total current flowing in the circuit.
.

Text Solution

Verified by Experts

(a) Since `R_1 and R_2` are in parallel, their resultant resistance `(R'_p)` is given by
`(1)/(R'_p) = (1)/(R_3) + (1)/(R_4) + (1)/(R_5)`
=`(1)/(30) + (1)/(20) + (1)/(60) = (6)/(60) = (1)/(10)`
or `R'_p = 10 Omega`
As `R_p and R'_p` are in series, total resistance in the circuit, `R_s = R_p + R'_p = 8 Omega + 10 Omega = 18 Omega`
(b) Total current in the circuit, i.e., `I = (V)/(R_s) = (12 V)/(18 Omega) = 0.67 A`.
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