Question

In the given circuit, the potential difference beween $\text{a}$ & $\text{b}$ is $300V$. The resistors are of $100\Omega$ each.

• A. If the upper meter $M_1$is an ideal voltmeter & lower meter $M_2$ is an ideal ammeter, reading of voltmeter is $300V$
• B. If both meters are ideal voltmeters, they show $200V$ each.
• C. If both meters are ideal ammeters, they read $6A$ each.
• D. If the upper meter $M_1$ is an ideal voltmeter and lower meter $M_2$ is an ideal ammeter, reading of ammeter is $3A$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A If the upper meter $M_1$is an ideal voltmeter & lower meter $M_2$ is an ideal ammeter, reading of voltmeter is $300V$
B If both meters are ideal voltmeters, they show $200V$ each.
C If both meters are ideal ammeters, they read $6A$ each.
D If the upper meter $M_1$ is an ideal voltmeter and lower meter $M_2$ is an ideal ammeter, reading of ammeter is $3A$.

For an ideal voltmeter, resistance $R_V=\infty$
For an ideal ammeter, resistance $R_A=0$

(A) Because resistance of an ideal ammeter is zero, all the current flows through the ammeter. Thus potential difference across ammeter is zero.


Reading of voltmeter = $300V$

(B)
Because $R_{V_1}=R_{V_2}=\infty$, current doesn't flow through either voltmeter.
$\therefore i=\frac{300V}{300\Omega }=1A$
$V_1=100\times 1+100\times 1=200V$
$V_2=100\times 1+100\times 1=200V$

(C) Because resistance of ideal ammeters is zero, potential differences across $A_1$ and $A_2$ are zero.

The resistances are in parallel combination.


$\therefore i=\frac{300V}{100/3\Omega }=9A$


Current through each resistor $=\frac{9}{3}=3A$
$\therefore$ Reading of ammeter is $6A$

(D)

Potential difference across $A_2$ is zero and all the current flows through $A_2$.

$\therefore i=\frac{300}{100}=3A$