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I am doing some practice questions. However, there doesn't appear to be any answers online. Therefore, I am checking if I am correct here. Could you please let me know if my analysis and calculations are correct.

The last Resistor R8 is short. Therefore, it is ignored and current will go through the short instead. R6 & R7 are in parallel 12||4 = 4 Ohms and R4 & R5 are in parallel too. 4 || 4 = 2 Ohms.

The calculated Ohms of those two parallel branches are added to R3 since they're now in series. Therefore, (3 + 4 + 2)Ohms = 9 Ohms. This is now in parallel with R2 which is 18 Ohms. 9 || 18 = 6 Ohms, plus R1 of 6 Ohms. Total resistance is 12 Ohms.

Source current = 54/12 = 4.5A

Current across I8 and V8 is 0 since this is short.


1 Answer 1


You're right that \$I_8=0A\$ and \$V_8=0V\$, for the right reasons.

You're right about the combined resistance of R3, R4, R5, R6, R7 and R8 being 9Ω.

Then you went wrong.

The 9Ω from before is in parallel with 18Ω of R2. That makes 6Ω.

Those 6Ω are in series with R1, for a total of 12Ω.

Those 12Ω are across the voltage source E, of 54V, so the final current is:

$$ I_S = \frac{54V}{12\Omega} = 4.5A $$

You got the right answer, 4.5A, with the wrong numbers.


I see you corrected the question. Now everything seems in order.

A little simulation for confirmation:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you for the help \$\endgroup\$ Aug 14, 2023 at 12:57

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