Question
Download Solution PDFConsider two circuits, A and B, each of which has 12 resistors each of resistance RA and RB. respectively. In each circuit the resistors are in such a way that the net resistance of each circuit is the minimum. Now a 30 V battery with negligible internal resistance is connected across each circuit separately, and the current drawn by circuit A and circuit B are 10 A and 18 A, respectively. Then RA and RB will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is option 4) RA = 36 Ω and RB = 20 Ω
Given:
12 resistors each of resistance RA and RB and it is having \(Ia=10A,Ib=18A\)
whose voltage is \(V=30V\)
Formula used:
Ohm's law: \(V=IR\)
Calculations:
As per the given statement in each circuit, the resistors are in such a way that the net resistance of each circuit is the minimum the resistors must be arranged in parallel. So their net resistance would be:
\({1 \over R'a}=12*{1 \over Ra}\)
\({1 \over R'b}=12*{1 \over Rb}\)
Thus using Ohm's law we get for circuit A:
\(V=IR\)
\(30=Ia*R'a\)
\(30=10*{Ra\over12}\)
\(Ra={12*30 \over 10}={360 \over 10}=36Ω\)
Similarly for circuit B we have:
\(V=IR\)
\(30=Ib*R'b\)
\(30=18*{Rb\over12}\)
\(Rb={12*30 \over 18}={360 \over 18}=20Ω\)
Therefore we have the resistances values as RA = 36 Ω and RB = 20 Ω.
Last updated on Jul 18, 2025
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