Question

When switch $S$ is closed, then the reading of ammeter is $A$, then value of $10A$ is ______

Answer 1

2.5 A

Answer 2

We first “read” the circuit as follows. The four resistors (20 Ω, 15 Ω, 25 Ω and 10 Ω) are arranged so that the 20 Ω, 15 Ω and 25 Ω resistors are in series (their net resistance is

$$R_{123}=20+15+25=60\;Ω$$

), and this branch is connected in parallel with the 10 Ω resistor. Their net (parallel) resistance is

$$R_{parallel}=\frac{10\times 60}{10+60}=\frac{600}{70}=\frac{60}{7}\;Ω.$$

Then, the 50 Ω resistor is connected in parallel with the combination whose equivalent is $60/7\;Ω$. However, note that the ammeter is inserted in series with the 25 Ω resistor which is part of the series branch (20 Ω, 15 Ω, 25 Ω). Since in a parallel network each branch sees the full applied voltage (15 V), the branch that contains the series combination (20+15+25 Ω) gets 15 V. Hence the current in that branch is

$$I_{\text{series}}=\frac{15}{60}=\;0.25\;A.$$

This current (0.25 A) is exactly what the ammeter reads (current through the 25 Ω resistor is the same as that in the entire series branch). Thus we have

$$A=0.25\;A.$$

Then

$$10A=10\times 0.25\;A=2.5\;A.$$

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Summary of the Minimal Core Solution:

1. Compute the series resistance of 20, 15, 25 Ω: $60\;Ω$.
2. Since $10\;Ω$ is in parallel with $60\;Ω$ the branch voltage is 15 V.
3. The series branch current (measured by the ammeter in series with the 25 Ω resistor) is $15/60=0.25\;A$.
4. Hence, $10A=10\times 0.25=2.5\;A$.