An electricity and magnetism problem by Bhavik Gohil

Electricity and Magnetism Level 2

In the circuit shown in fig, for R=20 ohms the current I is 2A. When R is 10 ohms the current I would be

2A 4A 1A 3A

2 solutions

Chew-Seong Cheong
Aug 17, 2014

The current through $R$ is always $4A$ because it is in series with the current source. Since $I = 2A$ , the current through the $20\Omega$ resistor is also $2A(=4-2)$ . This means that network $N_2$ has an equivalent resistance similar to the $20\Omega$ resistor. The constant $4A$ current through $R$ is split into two equal resistance of $20\Omega$ therefore, $I$ is always $2A$ .

@Chew-Seong Cheong this might be stupid, but i thought that since resistance*current is voltage, the current would be proportional to the resistance so i answered 1A

Mardokay Mosazghi - 6 years, 7 months ago

Mardokay, you are right that Ohm's law always applies but voltage division is not in this case because it involves a current source and not a voltage source. For a voltage source, its output voltage is always constant, the output current varies and we can use voltage division. For a current source, the output current is constant (4A), the voltage varies and we use current division as I have done in this problem.

The current through R in this case is always 4A. When R = 20 $\Omega$ , the voltage across R is 80V, when R = 10 $\Omega$ , it is 40V. Since the internal resistance of N $_2$ is 20 $\Omega$ , the 4A current is divided equally into 2A. The voltage across the 20 $\Omega$ resistor is always 40V,

I hope this is helpful.

Chew-Seong Cheong - 6 years, 7 months ago

Bhavik Gohil
Jun 20, 2014

AS current source is there in series with R, current through this will be irrespective of the value of resistance hence it will be 2A only

An electricity and magnetism problem by Bhavik Gohil

2 solutions

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