Question

In figure the voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are:

• A. 32.978 V, 3 A
• B. 3 V, 0.236 A
• C. 120 V, 4.198 A
• D. 120 V, 3 A

Answer 1

The correct option is B 3 V, 0.236 A

$\to$ Ideal voltmeter means it has $\infty$ resistance, current will not flow in the branch of a voltmeter.

$\to$ Ideal ammeter means it has zero resistance.So that it can read maximum current.

Given $V_A-V_C=6V\to$ (1)

$V_B-V_C=1.50V\to$ (2)

Subtract equation $\left(1\right)$ and $\left(2\right)$

$V_A-V_B=4.5V$

So reading of voltmeter $=V_A-V_B=4.5V$

Now we can assume that $EMF6V$ has an internal resistance of $10\Omega$ and $1.5V$ has $5\Omega$.

Both are in parallel combination.

$\therefore E_{eq}=\frac{E_1{r}_2=E_2{r}_1}{r_1+r_2}$

$=\frac{6\times 5+1.5\times 10}{10+5}$

$=\frac{30+15}{15}$

$=3V$

$r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

$=\frac{10\times 5}{15}$

$=\frac{50}{15}=\frac{10}{3}\Omega$

Now the circuit will becomes

Equal current will flow through the same resistance.

Using Kirchhoff's law in loop $\left(1\right)$

$\frac{i}{2}\times 6-3+\frac{10}{3}\times i=0$

$\frac{6i}{2}+\frac{10i}{3}=3$

$(3+\frac{10}{3})i=3$

$\frac{19}{3}i=3$

$[i=\frac{9}{19}A]$

Ammeter will read the reading$=\frac{i}{2}$

$=\frac{9}{19\times 2}$

$=\frac{9}{38}A=0.236A$

Hence B is correct option.