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A linear circuit is shown in the figure.

The elements in this circuit have the following values: R = 100 Ohms, R2 = 200 Ohms, R3 = 350 Ohms, V = 5V, and I = 0.004 A.

  1. Determine the potentials of drops v_a and v_b across resistors R1 and R2 respectively.

  2. Now let us determine if the answers you came up with satisfy the laws of physics. (a) What is the power (in Watts) dissipated in resistor R1? (b) What is the power (in Watts) dissipated in resistor R2? (c) What is the power (in Watts) dissipated in resistor R3? (d) What is the power (in Watts) coming out of the voltage source V? (e) What is the power (in Watts) coming out of the current source I?

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From the circuit I have:

\$-v_b + v_a + V = 0\$

\$v_b - V = v_a\$

\$i_1 = (v_b - V)/R_1\$

\$I + i_2 = i_1\$

\$(v_b - V)/R_1 = I + v_b/R_2\$

From this last equation I get \$v_b = 10.8\$ and hence \$v_a = 5.8\$.

However, apparently that is wrong. (And hence my answers to #2 were all wrong as well.) Why is that so? What might I be doing wrong?

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  • \$\begingroup\$ Assuming \$R=R_1\$, there is only one unknown node voltage, which is \$V_\text{B}\$. (Just ground the bottom node to make it \$0\:\text{V}\$.) The result should be \$V_\text{B}=3.6\:\text{V}\$. From there, the answers just flow out. Do you see how to develop that voltage? \$\endgroup\$ – jonk Jan 13 at 7:54
  • \$\begingroup\$ Wrong sign in the last equation. And you haven't stated where i2 is. \$\endgroup\$ – Chu Jan 13 at 12:50
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I believe your problem is with the 4th equation. As defined in the circuit, all of the currents are entering the node so their sum must be zero.

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  • \$\begingroup\$ But the currents are not defined in the circuit. There is no marking of \$i_1\$ or \$i_2\$ in the circuit. \$\endgroup\$ – Elliot Alderson Jan 13 at 18:31

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