DC Current Splitting

A D C DC voltage source V S V_S and a D C DC current source I S I_S feed a circuit with resistors R 1 R_1 and R 2 R_2 as shown. What is the current through R 2 R_2 ?

Details and Assumptions:
1) V S = 10 V_S = 10 volts
2) I S = 5 I_S = 5 amps
3) R 1 = 2 Ω R_1 = 2 \Omega
4) R 2 = 3 Ω R_2 = 3 \Omega
5) Give your answer as a positive number
6) Source polarities are as indicated in the diagram


The answer is 4.0.

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2 solutions

Tom Engelsman
May 11, 2021

Let V 0 V_{0} be the potential at the top-center node above resistor R 1 . R_{1}. Applying KCL at this node yields the relationship:

I S = I R 1 + I R 2 = V 0 V S R 1 + V 0 R 2 \Large I_{S} = I_{R1} + I_{R2} = \frac{V_{0}-V_{S}}{R_{1}} + \frac{V_{0}}{R_{2}}

and solving for V 0 V_{0} produces:

( I S + V S R 1 ) ( R 1 R 2 R 1 + R 2 ) = V 0 \Large (I_{S} + \frac{V_{S}}{R_{1}})(\frac{R_{1}R_{2}}{R_{1}+R_{2}}) = V_{0} ;

or ( 5 + 10 2 ) ( 2 3 2 + 3 ) = V 0 ; \Large (5 + \frac{10}{2})(\frac{2 \cdot 3}{2+3}) = V_{0};

or V 0 = 12 V_{0} =12 volts. Finally, I R 2 = V 0 R 2 = 12 3 = 4 \large I_{R2} = \frac{V_{0}}{R_{2}} = \frac{12}{3} = \boxed{4} amps.

Veeresh Pandey
May 13, 2021

Use current division and superposition

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