Find the current in the 12-Ω resistor in the figure below. (Assume R1 = R3 = 2.8 Ω, R2 = R4 = 7.8 Ω, ΔV = 21 V.)
Consider I is the current through the circuit. This same current
flows through R3 and the it branches into two. One flowing in the
upper network and another in the lower network.
Resistance of the upper network
Ru = R1 + (R2||R4)
Here "+" is used for series connection and "||" is used for
parallel connection.
Ru = R1 + [(R2 R4) / (R2 + R4)]
Ru = 2.8 + [7.8 x 7.8 / (7.8 + 7.8)]
= 2.8 + 3.9 = 6.7 Ω
Resistance in the lower network,
Rl = (4||12) + 2
= [(4 x 12) / (4 + 12)] + 2
= 5 Ω
Resistance of the entire circuit, R = R3 + (Ru||Rl)
= 2.8 + [(6.7 x 5) / (6.7 + 5)]
= 5.66 Ω
Current in the curcuit = I = ΔV/R
= 21/5.66 = 3.71 A
Current in the lower circuit, Il = I x Ru /
(Ru+Rl)
= 3.71 x [6.7 / (6.7 + 5)]
= 2.12 A.
Current in the 12 Ω resistor
= Il x 4 / (4 + 12)
= 0.53 A
Find the current in the 12-Ω resistor in the figure below. (Assume R1 = R3 =...
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