for the circuit shown in the figure the ammeter a2 reads 1.6 a
For this well-known alternating current (AC) question, the key idea is that the different branch currents are out of phase, so you cannot simply add them arithmetically.
Core idea of the problem
The typical setup is:
- Ammeter A₂ reads 1.6 A (current in one reactive branch, say inductive), which lags the supply e.m.f. by π/2\pi/2π/2.
- Ammeter A₃ reads 0.4 A (current in another reactive branch, say capacitive), which leads the supply e.m.f. by π/2\pi/2π/2.
Because one current lags by π/2\pi/2π/2 and the other leads by π/2\pi/2π/2, these two branch currents are 180° out of phase with each other.
So their phasor (vector) sum is:
I1=1.6textA−0.4textA=1.2textAI_1=1.6\\text{A}-0.4\\text{A}=1.2\\text{A}I1=1.6textA−0.4textA=1.2textA
Hence the ammeter A₁ (measuring the net current) reads 1.2 A.
In words: the larger branch current (1.6 A) and the smaller one (0.4 A) oppose each other in phase, so the net is their difference, not their sum.
What this tells you physically
- The 1.6 A branch behaves like a pure inductor (current lagging voltage by 90°).
- The 0.4 A branch behaves like a pure capacitor (current leading voltage by 90°).
- Because they are equal-frequency, equal-voltage branches in series/parallel AC , their reactances relate as
XCXL=1.60.4=4,\frac{X_C}{X_L}=\frac{1.6}{0.4}=4,XLXC=0.41.6=4,
which leads (in the full solution) to
ω=12LC,f=14πLC.[]\omega =\frac{1}{2\sqrt{LC}},\quad f=\frac{1}{4\pi\sqrt{LC}}.[]ω=2LC1,f=4πLC1.[]
So for the classic question “for the circuit shown in the figure, the ammeter A₂ reads 1.6 A and A₃ reads 0.4 A, then what does A₁ read?” the correct reading is:
A₁ = 1.2 A.
If your version only states “A₂ reads 1.6 A” but not “A₃ reads 0.4 A,” it is almost certainly this same standard problem whose full statement includes both readings.