A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per mill

Q: Why do we divide total divisions by sensitivity?

Sensitivity is given as divisions per unit (mA or mV). To find the total units required for a full-scale deflection, we divide the total number of divisions by the divisions per unit.

Q: Why is the series resistance so high?

A voltmeter must have a very high resistance so that it draws negligible current from the circuit it is measuring. This ensures that connecting the voltmeter doesn't alter the actual voltage you are trying to measure.

Q: What happens if we connect the resistance in parallel instead of series?

Connecting a low resistance in parallel (called a shunt) converts the galvanometer into an ammeter, which is used to measure current, not voltage.

Q: Why is the new full-scale voltage exactly 150 V?

The problem states that we want each of the 150 divisions to read 1 V. Therefore, the maximum reading (when the needle is at the very end of the scale) is $150 \times 1 = 150 \text{ V}$.

Q: Can we use any galvanometer to measure 150 V?

Yes, theoretically, as long as we connect an appropriately calculated high resistance in series. This resistance ensures that even at 150 V, the current flowing through the galvanometer does not exceed its safe full-scale value ($I_g$).

Question

A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V, the resistance in Ohms needed to be connected in series with the coil will be

Options:

(A) $10^3$

(B) $10^5$

(C) 99995

(D) 9995

Sigma Percentile · Accepted Answer

### Decoding the Galvanometer's DNA

Imagine a galvanometer as a highly sensitive instrument, capable of detecting even the faintest whispers of electrical current. But to use it effectively, we first need to understand its intrinsic properties. The problem gives us two crucial pieces of information: **current sensitivity** and **voltage sensitivity**.

Current sensitivity tells us how many divisions the needle deflects for every milliampere of current. Since there are $150$ divisions in total, the maximum current the galvanometer can handle—its full-scale deflection current ($I_g$)—is simply the total divisions divided by the current sensitivity:

$$I_g = \frac{150}{10} = 15 	ext{ mA}$$

Similarly, voltage sensitivity tells us the deflection per millivolt. The full-scale deflection voltage ($V_g$) is:

$$V_g = \frac{150}{2} = 75 	ext{ mV}$$

### Unveiling the Internal Resistance

Every physical coil has some resistance. Now that we know the maximum voltage and current, Ohm's law comes to our rescue to find the galvanometer's internal resistance ($G$).

$$G = \frac{V_g}{I_g} = \frac{75 	ext{ mV}}{15 	ext{ mA}} = 5 \Omega$$

This $5 \Omega$ is the inherent resistance of the galvanometer coil itself.

### The Voltmeter Transformation

Our goal is to transform this delicate instrument into a robust voltmeter where each of its $150$ divisions represents $1 	ext{ V}$. This means the new full-scale voltage ($V$) we want to measure is:

$$V = 150 	imes 1 = 150 	ext{ V}$$

To achieve this without frying the delicate coil (which can only handle $75 	ext{ mV}$), we must connect a large resistance ($R$) in series. This series resistance acts as a shield, dropping most of the voltage across itself and allowing only a safe fraction to reach the galvanometer.

The master equation for this conversion is:

$$V = I_g(R + G)$$

### The Final Calculation

Now, it's just a matter of plugging in our known values and solving for $R$. Let's substitute the values we've found:

$$150 = 15 	imes 10^{-3} (R + 5)$$

Dividing both sides by $15 	imes 10^{-3}$ gives us:

$$10^4 = R + 5$$

$$10000 = R + 5$$

$$R = 9995 \Omega$$

And there we have it! By connecting a $9995 \Omega$ resistor in series, we successfully upgrade our galvanometer into a $150 	ext{ V}$ voltmeter.

Select Answer:

Visualized Solution

$I_{g} = \frac{Total Divisions}{Current Sensitivity}$

$V_{g} = \frac{Total Divisions}{Voltage Sensitivity}$

$G = \frac{V _{g}}{I _{g}}$

$New Voltage per division = 1 V$

$V = I_{g} (R + G)$

$1 0^{4} = R + 5$

The Sigma Insight: Moving Coil Galvanometer

Decoding the Galvanometer's DNA

Unveiling the Internal Resistance

The Voltmeter Transformation

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ig​=Current SensitivityTotal Divisions​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vg​=Voltage SensitivityTotal Divisions​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G=Ig​Vg​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Voltage per division=1 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=Ig​(R+G)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 104=R+5