Q.In the circuit (Fig. 4.23) the current is to be measured. What is the value of the current if the ammeter shown

The key idea is ammeter loading effect: a real ammeter has non-zero resistance that alters the circuit current, while an ideal ammeter has zero resistance and reads the true current.

Reasoning:

1. The circuit is a single loop with a $3.00\ \text{V}$ battery and a $3.00\ \Omega$ resistor. The ammeter is in series.
2. For each case, total resistance $R_{\text{total}} = 3.00\ \Omega + R_{\text{ammeter}}$. Current $I = \frac{3.00\ \text{V}}{R_{\text{total}}}$.
3. (a) $R_{\text{ammeter}} = R_G = 60.00\ \Omega$ → $R_{\text{total}} = 63.00\ \Omega$ → $I = 3.00/63.00 = 0.04762\ \text{A}$. …

The core idea here is the ammeter loading effect. An ammeter must be placed in series with the component whose current you want to measure. That means the ammeter's own resistance adds directly to the circuit's total resistance. If the ammeter's resistance is not negligible compared to the rest of the circuit, it reduces the current below what it would be in the undisturbed circuit. The "true" current — the one we actually want to measure — is the current that flows when the ammeter has zero resistance (an ideal ammeter).

The circuit is a simple single-loop: a 3.00 V battery connected in series with a 3.00 Ω resistor and the ammeter. The ammeter's resistance changes in each part of the question.

1. Find the true (ideal) current first. This gives us a reference point. With an ideal ammeter ($R_A = 0\ \Omega$), the total resistance is just the 3.00 Ω resistor.

This is the current that would flow if the ammeter didn't interfere at all.

2. Part (a): Galvanometer alone. A galvanometer is a sensitive current detector, but it has a significant internal resistance — here $R_G = 60.00\ \Omega$. When placed directly in the circuit as an ammeter, the total resistance becomes:

The current is then:

This is drastically smaller than 1 A. The galvanometer's high resistance has "loaded" the circuit, giving a completely misleading reading. This is why you never use a bare galvanometer as an ammeter for anything but tiny currents.

3. Part (b): Galvanometer with a shunt.

   To convert a galvanometer into a useful ammeter, we connect a very small resistance $r_s = 0.02\ \Omega$ in parallel with the galvanometer coil. This shunt provides an alternative path for most of the current, protecting the delicate galvanometer and drastically reducing the overall resistance of the ammeter.

   The ammeter's effective resistance $R_A$ is the parallel combination of $R_G$ and $r_s$: …

Method: Solving Ammeter-Loading / Meter-Conversion Circuits

Any question that asks "what current does a real meter read compared to the true current" reduces to one repeated calculation: find the meter's effective series resistance, add it to the rest of the circuit, and re-apply Ohm's law.

Steps

Step 1: Find the true (ideal-meter) current as a reference

Set the meter's resistance to zero and solve the bare circuit with Ohm's law:

This is the benchmark every other case is compared against -- it tells you how far "loading" pushes the reading away from reality.

Step 2: Find the meter's own effective resistance for the given configuration

• A bare galvanometer used directly as an ammeter contributes its full coil resistance $R_G$.
• A galvanometer converted to an ammeter with a shunt $r_s$ in parallel contributes the parallel combination:

Because $r_s \ll R_G$ in a working shunt, $R_{\text{meter}} \approx r_s$ -- the shunt, not the galvanometer coil, dominates.

Step 3: Add it in series and recompute the current …