In the below circuit, \$ A_1, A_2 \text{ and } A_3 \$ are ideal ammeters.
If \$A_2\$ and \$A_3\$ read \$3A\$ and \$4A\$ respectively, then \$A_1\$ should read ______?


My Approach:
Current in resistor \$ R_1 \$ is defined as \$ 3A \$
Current in resistor \$R_2 \$ is defined as \$ 4A \$
Therefore, \$ I_1 = 3A + 4A = 7A \$

Since we have sinusoidal source, therefore ammeter reads r.m.s value of current
Therefore, \$ I_1= \sqrt{3^2 + 4^2} = 5A \$

so which is the correct answer? please help...

  • \$\begingroup\$ This is no "approach". That's just presenting two answers without real justification for any of them. \$\endgroup\$
    – Curd
    Mar 22, 2018 at 7:56
  • 4
    \$\begingroup\$ AC current doesn't magically require you to calculate geometric sums. Use the same Kirchhoff rules you learned for DC. \$\endgroup\$
    – Janka
    Mar 22, 2018 at 8:00
  • \$\begingroup\$ Why are you saying so? @Curd ...I have shown how I am getting the two answers, given the reasons before writing the two answers and hence it is my approach ; if i just wrote \$ 7A \$ and \$ 5A \$ as its answers then you can say its no approach \$\endgroup\$
    – Suresh
    Mar 22, 2018 at 8:01
  • \$\begingroup\$ @Suresh: ok, you showed how you got the answers. That's right; but I think the point in the problem is not to show that you can do a normal addition (3+4=7) or a vector addition (vector of length 3 + vector of length 4 perpendicular to first vector = vector of length 5). It's to show that you know when/why to use which of the two; at least some hints (no matter if wrong or not); but that's missing. \$\endgroup\$
    – Curd
    Mar 22, 2018 at 8:13

2 Answers 2


The current that goes through A1 has to split at every instance of time between paths A2 and A3. It doesn't matter if it is AC or DC. At any instance of time Kirchhoff's rule must apply.

So if we imagine it is DC then obviously A1=A2+A3. However all that is happening with AC is that the source voltage is rising and falling, but A1 at all times must still be A2+A3.

To read current in an AC circuit we might require the ammeter to read an RMS value. This just means that the meter integrates the value of I^2 over a long period of time relative to the frequency of the source and displays the square root of this value. Maybe by measuring the heating effect in a small piece of wire.

This is the same for all the meters. This process of RMS calculation actually makes no difference because all the meters do it independently. If you imagine the source was DC then these meters would each display the square root of I^2 for each, which is just I. So if you start to increase the frequency it doesn't change anything about the calculation that A1=A2+A3.

  • 1
    \$\begingroup\$ The key here is that there are no reactive components (i.e., capacitors or inductors) in the circuit. So AC acts just like DC, because resistors don't care about the AC fluctuations. \$\endgroup\$ Mar 22, 2018 at 11:37
  • \$\begingroup\$ @PeteBecker This is a very very good point that I forgot to mention in my response. If R1 or R2 had inductive or capacitative impedance, then the voltage and currents would depend on the supply frequency and would be obtained by vector calculations, because the A1 current may not be in phase with A2, or A3. For example if R1 was an inductor and R2 was a capacitor then the instantaneous current vectors at A2 and A3 would point in the opposite direction, or if one was a resistor then they would be out of phase by 90º. \$\endgroup\$
    – Robotbugs
    Mar 24, 2018 at 8:18
  • \$\begingroup\$ Also note that since the circuit only uses passive linear components, even if they were R, C, or Ls, the current waveforms will still only be sine waves at the generator frequency, and this means that the peak currents will always be 1.414 times the RMS currents. This scale factor applies to all RMS measurements of continuous sine wave signals. The scale factor would be different if the wave was half or full wave rectified. \$\endgroup\$
    – Robotbugs
    Mar 24, 2018 at 8:24

At any instant , KCL will hold true whether it's AC or DC circuit. If it's a sine wave source and all your ammeters are reading RMS values, then A2 and A3 are reading 3 and 4 Amperes as the RMS values of the respective currents. You just have to add them to find the reading on A1, which is also RMS value.


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