A constant voltage of $50\text{ V}$ is maintained between the points A and B of the circuit shown in the figure (bridge network with resistors $1\,\Omega$, $2\,\Omega$, $3\,\Omega$, $4\,\Omega$ and central vertical branch CD). The current through the branch CD of the circuit is:

A 1.5 A

B 2.0 A

C 2.5 A

D 3.0 A

The answer is B.

Solution

Applying Kirchhoff's loop laws for the circuit under $50\text{ V}$ bias yields current $i_{CD} = 2.0\text{ A}$.

Using Kirchhoff's loop equations: For loop ACDEA: $i_1(1) - i_2(3) = 0 \implies i_1 = 3i_2$. For loop CBFDC: $3 i_3 + 2 i_2 - i_1 = 0 \implies i_2 = 3 i_3$. For outer loop EDFGH with $50\text{ V}$ battery: $-3 i_2 - 4(i_2 + i_3) + 50 = 0 \implies 7 i_2 + 4 i_3 = 50$. Substituting $i_2 = 3 i_3$: $7(3 i_3) + 4 i_3 = 50 \implies 25 i_3 = 50 \implies i_3 = 2.0\text{ A}$. Thus, the current through branch CD is $2.0\text{ A}$.

Concept Kirchhoff's Voltage Law

Current Electricity in past papers

130 questions from this chapter have appeared across 27 exam years.

A constant voltage of $50\text{ V}$ is maintained between the points A and B of the circuit shown in the figure (bridge network with resistors $1\,\Omega$, $2\,\Omega$, $3\,\Omega$, $4\,\Omega$ and central vertical branch CD). The current through the branch CD of the circuit is:

Solution

Current Electricity in past papers

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