A galvanometer has full scale deflection current of 500 micro-Ampere, and resistance of 100 Ohm. Shunt with what resistance has to be combined with the galvanometer in order to transform it to an ammeter able to measure current of 10 Ampere full scale?

 To convert a galvanometer into an ammeter capable of measuring higher currents, we need to connect a low resistance in parallel with the galvanometer. This low resistance is called a "shunt" resistance, and it allows most of the current to bypass the galvanometer to prevent it from being damaged by high currents.

Let's calculate the shunt resistance required for the given conditions:

1. Full-scale deflection current of the galvanometer, \( I_g \), is 500 microamperes (µA) or 500 x 10^(-6) A. 2. Galvanometer resistance, \( R_g \), is 100 ohms (Ω). 3. The desired full-scale deflection current of the converted ammeter, \( I_A \), is 10 amperes (A).

Since we want the galvanometer to deflect fully for a current of 10 A, the shunt resistance, \( R_s \), will carry the rest of the current which is not going through the galvanometer. Let's denote the current through the shunt as \( I_s \).

Kirchhoff's Current Law states that the total current entering a junction must equal the total current leaving the junction. Applying this to our setup we get: \[ I_A = I_g + I_s \]

Rearranging to find the current through the shunt: \[ I_s = I_A - I_g \] \[ I_s = 10 A - 500 x 10^(-6) A \] \[ I_s \approx 10 A \] (since 500 µA is negligible compared to 10 A)

Now, the voltage across the shunt (\( V_s \)) and the galvanometer (\( V_g \)) will be the same due to parallel connection. We apply Ohm's law to find the voltage across the galvanometer: \[ V_g = I_g \cdot R_g \] \[ V_g = 500 x 10^(-6) A \cdot 100 Ω \] \[ V_g = 0.05 V \]

By Ohm's law, the voltage across the shunt is also 0.05 V. We can now find the shunt resistance using Ohm's law (\( V = I \cdot R \)): \[ R_s = \frac{V_s}{I_s} \] \[ R_s = \frac{0.05 V}{10 A} \] \[ R_s = 0.005 Ω \] or 5 milliohms (mΩ)

Therefore, to transform the given galvanometer into an ammeter that can measure up to 10 amperes, we need to connect a shunt resistance of 5 milliohms in parallel with it.