Question: 36. A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
(1) 1.5 A
(2) 2.0 A
(3) 2.5 A
(4) 3.0 A
Answer: Option (2)
Explanation:
The branch CD is just a connecting wire,
so points C and D are at the same potential.
Let this common potential be V_X.
Take the potential at A as 50 V
and at B as 0 V.
The resistors between A and node X
(i.e. C and D) are 1\ \Omega and 3\ \Omega, both connected from A(50\ \text{V}) to V_X.
Current through 1\ \Omega resistor: I_1 = \dfrac{50 - V_X}{1}
Current through 3\ \Omega resistor: I_3 = \dfrac{50 - V_X}{3}
The resistors between node X and B are 2\ \Omega and 4\ \Omega, both connected from V_X to 0 V.
Current through 2\ \Omega resistor: I_2 = \dfrac{V_X}{2}
Current through 4\ \Omega resistor: I_4 = \dfrac{V_X}{4}
Apply Kirchhoff’s current law at node X: total current coming from A equals total current going to B.
\dfrac{50 - V_X}{1} + \dfrac{50 - V_X}{3} = \dfrac{V_X}{2} + \dfrac{V_X}{4}Simplify:
(50 - V_X)\left(1 + \dfrac{1}{3}\right) = V_X\left(\dfrac{1}{2} + \dfrac{1}{4}\right) (50 - V_X)\dfrac{4}{3} = V_X\dfrac{3}{4}Multiply both sides by 12:
16(50 - V_X) = 9V_X 800 - 16V_X = 9V_X 800 = 25V_X \Rightarrow V_X = \dfrac{800}{25} = 32\ \text{V}Now find currents in the top branch to get current in CD.
Current through 1\ \Omega: I_1 = \dfrac{50 - 32}{1} = 18\ \text{A}
Current through 2\ \Omega: I_2 = \dfrac{32}{2} = 16\ \text{A}
At point C, 18 A comes in from the 1\ \Omega resistor.
Out of this, 16 A goes to B through the 2\ \Omega resistor,
and the remaining current must flow through branch CD.
So current through branch CD: