Consider the circuits shown in the figures. Both the circuits are taking the same current from battery but current through R in the second circuit is 110th of current through R in the first circuit. If R is 11Ω the value of R1(inΩ) is
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Solution
In first figure, from ohm's law i=E/R
In second figure by KCL at junction A, current through
R2=i−i10=9i10
Potential difference across R2 = Potential difference across R (parallel combination)
⇒R2×910i=R×i10
R2=R9=119Ω
for R2 and R connected in parallel, equivalent resistance
Req=R2×R(R2+R)=119×111119+111=1110Ω
R1 is in series with above combination
for second figure, total circuit resistance