Question

A perfect voltmeter is connected between C & D gives reading 140 V. When we replace it with an ideal ammeter, its reading is 21 A. Now, we connect a resistor r between C & D. If the current I (in A) is flowing through this r between C & D. Find the value of $\frac{I}{4}$.

Answer 1

2

Answer 2

The circuit is a Wheatstone bridge with resistors $R$ in arms AC and DB, and resistors $r$ in arms CB and AD. A battery with emf $\mathcal{E} = 420$ V is connected between A and B.

Case 1: Perfect voltmeter between C and D.

A perfect voltmeter has infinite resistance, so no current flows through it. The potential difference between C and D is measured.

The voltage at C is $V_C = V_A - I_{AC} R$, where $I_{AC}$ is the current in arm AC. The current in the upper branch ACB is $I_1 = \frac{\mathcal{E}}{R+r}$. So $V_C = V_A - \frac{\mathcal{E}}{R+r} R$.

The voltage at D is $V_D = V_A - I_{AD} r$, where $I_{AD}$ is the current in arm AD. The current in the lower branch ADB is $I_2 = \frac{\mathcal{E}}{r+R}$. So $V_D = V_A - \frac{\mathcal{E}}{r+R} r$.

The voltmeter reading is $V_{CD} = V_C - V_D = \left(V_A - \frac{\mathcal{E}R}{R+r}\right) - \left(V_A - \frac{\mathcal{E}r}{r+R}\right) = \mathcal{E} \left(\frac{r}{r+R} - \frac{R}{R+r}\right) = \mathcal{E} \frac{r-R}{R+r}$.

Given $V_{CD} = 140$ V and $\mathcal{E} = 420$ V.

$140 = 420 \frac{r-R}{R+r} \implies \frac{r-R}{R+r} = \frac{140}{420} = \frac{1}{3}$.

$3(r-R) = R+r \implies 3r - 3R = R+r \implies 2r = 4R \implies r = 2R$.

Case 2: Ideal ammeter between C and D.

An ideal ammeter has zero resistance, so C and D are short-circuited. The current through the ammeter is measured.

When C and D are shorted, the circuit becomes R in parallel with r, in series with r in parallel with R.

Equivalent resistance $R_{eq} = \frac{Rr}{R+r} + \frac{rR}{r+R} = \frac{2Rr}{R+r}$.

Total current from battery $I_{total} = \frac{\mathcal{E}}{R_{eq}} = \frac{420}{\frac{2Rr}{R+r}} = \frac{420(R+r)}{2Rr} = \frac{210(R+r)}{Rr}$.

Current through ammeter $I_{CD}$ can be found using current division or nodal analysis.

Let's use nodal analysis. Let $V_C = V_D = V$. Let $V_A = 420$ V and $V_B = 0$ V.

Current from A to C is $I_{AC} = \frac{V_A - V}{R}$. Current from A to D is $I_{AD} = \frac{V_A - V}{r}$.

Current from C to B is $I_{CB} = \frac{V - V_B}{r}$. Current from D to B is $I_{DB} = \frac{V - V_B}{R}$.

Current through ammeter from C to D is $I_{CD}$.

At node C: $I_{AC} = I_{CB} + I_{CD} \implies \frac{420-V}{R} = \frac{V-0}{r} + I_{CD}$.

At node D: $I_{AD} + I_{CD} = I_{DB} \implies \frac{420-V}{r} + I_{CD} = \frac{V-0}{R}$.

Subtracting the two equations: $\frac{420-V}{R} - \frac{420-V}{r} = \frac{V}{r} - \frac{V}{R}$.

$(420-V)\left(\frac{1}{R} - \frac{1}{r}\right) = -V\left(\frac{1}{R} - \frac{1}{r}\right)$.

If $\frac{1}{R} - \frac{1}{r} \neq 0$, then $420-V = -V \implies 420 = 0$, which is impossible.

So, $\frac{1}{R} - \frac{1}{r} = 0 \implies R=r$. But we know $r=2R$.

Let's recheck the setup. The ammeter current is the current flowing from C to D.

$I_{CD} = I_{AC} - I_{CB} = \frac{V_A - V}{R} - \frac{V - V_B}{r}$.

$I_{CD} = I_{DB} - I_{AD} = \frac{V - V_B}{R} - \frac{V_A - V}{r}$.

With $r=2R$:

$I_{CD} = \frac{420 - V}{R} - \frac{V}{2R} = \frac{840 - 2V - V}{2R} = \frac{840 - 3V}{2R}$.

$I_{CD} = \frac{V}{R} - \frac{420 - V}{2R} = \frac{2V - (420 - V)}{2R} = \frac{3V - 420}{2R}$.

$\frac{840 - 3V}{2R} = \frac{3V - 420}{2R} \implies 840 - 3V = 3V - 420 \implies 1260 = 6V \implies V = 210$ V.

This is the potential at C and D when shorted.

$I_{CD} = \frac{840 - 3(210)}{2R} = \frac{840 - 630}{2R} = \frac{210}{2R} = \frac{105}{R}$.

Given ammeter reading is 21 A.

$21 = \frac{105}{R} \implies R = \frac{105}{21} = 5 \, \Omega$.

Then $r = 2R = 2 \times 5 = 10 \, \Omega$.

Case 3: Resistor r is connected between C and D.

The resistance connected is $r = 10 \, \Omega$. We need to find the current I flowing through this resistor.

We can use the Thevenin equivalent circuit across C and D.

The Thevenin voltage $V_{Th} = V_{CD}$ (open circuit) = 140 V.

The Thevenin resistance $R_{Th}$ is the resistance between C and D with the battery short-circuited.

When the battery is short-circuited, A and B are connected.

The resistance between C and D is the parallel combination of the resistance through A and the resistance through B.

Resistance through A: R in series with r = R+r.

Resistance through B: r in series with R = r+R.

$R_{Th} = (R+r) || (r+R) = \frac{(R+r)(r+R)}{(R+r)+(r+R)} = \frac{(R+r)^2}{2(R+r)} = \frac{R+r}{2}$.

With $R=5 \, \Omega$ and $r=10 \, \Omega$, $R_{Th} = \frac{5+10}{2} = \frac{15}{2} = 7.5 \, \Omega$.

Now, the resistor r=10 $\Omega$ is connected between C and D. This resistor is connected across the Thevenin terminals.

The current I through this resistor is $I = \frac{V_{Th}}{R_{Th} + r} = \frac{140}{7.5 + 10} = \frac{140}{17.5}$.

$17.5 = \frac{35}{2}$.

$I = \frac{140}{35/2} = \frac{140 \times 2}{35} = \frac{4 \times 35 \times 2}{35} = 4 \times 2 = 8$ A.

The current flowing through the resistor r is I = 8 A.

We need to find the value of $\frac{I}{4}$.

$\frac{I}{4} = \frac{8}{4} = 2$.