Assume a network consisting of N number of resistors, dependent current sources and voltage sources also independent current sources and voltage sources (ideal). Now the current across one particular resistor is I Amps. To get 2I Amps which one of the following would be appropriate?

A] Double the independent voltage sources.  
B] Double the independent current sources.  
C] Double the values of dependent current sources and voltage sources.  
D] Double values of both independent current and voltage sources.  
E] Can't be determined.  

How the current relates to dependent or independent voltage source here?

  • 2
    \$\begingroup\$ Draw a simple circuit with one voltage source and, say, two resistors in series. Try to answer for that. Now draw another one with two resistors in parallel. Will the previous answer apply? Now draw another one, with one resistor in series with two in parallel. Will any of the previous answer apply? Now try changing the voltage source to current source and repeat. \$\endgroup\$
    – Eugene Sh.
    Apr 11 at 14:53
  • \$\begingroup\$ What to do with dependent sources then? This question was asked in previous years exams, isn't there any faster approach. \$\endgroup\$
    – Adyy
    Apr 11 at 15:07
  • 1
    \$\begingroup\$ The faster approach is to use your experience/intuition, which you may not have developed yet. Forget about dependent sources, it can be answered ignoring this part. \$\endgroup\$
    – Eugene Sh.
    Apr 11 at 15:08
  • \$\begingroup\$ BTW, are you supposed to select a single option or multiple? \$\endgroup\$
    – Eugene Sh.
    Apr 11 at 15:11
  • \$\begingroup\$ Only one of the options is correct \$\endgroup\$
    – Adyy
    Apr 11 at 15:22

1 Answer 1


Assuming we are talking about a linear circuit (meaning, the resistors and dependent sources are linear devices), then the answer is (D) --- double the value of all independent sources in the system.

To understand why, consider first that because of linearity the initial current I can be written as the sum of the currents through the chosen resistor due to each of the independent sources.

And second (again, because of linearity) that if any of the independent sources is modified by multiplying its value by some factor A, then the contribution of that source to the current through the resistor will also scale by A.


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