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Question

Physics Question on Electric Current

The current passing through the ideal ammeter in the circuit given below is

A

1.25 A

B

1 A

C

0.75 A

D

0.5 A

Answer

0.5 A

Explanation

Solution

Here, 2Ω2 \Omega and 2Ω2 \Omega are in parallel
1R=12+12\therefore \frac{1}{R}=\frac{1}{2}+\frac{1}{2}
R=2×22+2=1ΩR=\frac{2 \times 2}{2+2}= 1 \,\Omega
Now, internal resistance (1Ω),2Ω,4Ω(1\, \Omega), 2 \,\Omega, 4\, \Omega and resistance RR are in series
Rnet=1Ω+2Ω+4Ω+Ω=8Ω\therefore R_{n e t}=1 \Omega+2 \Omega+4 \Omega+\lfloor\Omega=8 \Omega
Hence, current I=VR=48=0.5AI=\frac{V}{R}=\frac{4}{8}=0.5 A