An ammeter of resistance 5Ω has a full-scale deflection when a current of 50 mA flows in it. The value of the resistor required to adapt it to measure a cur...

Answer Details

The ammeter has a resistance of 5Ω and reads full scale when a current of 50 mA flows through it. To measure a current of 5A, a shunt resistor is required to be connected in parallel with the ammeter. We can use the formula I(total) = I(ammeter) + I(shunt), where I(total) is the total current, I(ammeter) is the current passing through the ammeter, and I(shunt) is the current passing through the shunt resistor. Since the ammeter reads full scale at 50 mA, it means that 50 mA will flow through the ammeter when a total current of 5 A is passing through the circuit. Therefore, we can substitute the values in the formula: 5A = 50mA + I(shunt) Solving for I(shunt), we get: I(shunt) = 5A - 50mA I(shunt) = 4.95A To find the value of the shunt resistor, we can use Ohm's Law: V = IR Where V is the voltage drop across the shunt resistor, I is the current passing through the shunt resistor, and R is the resistance of the shunt resistor. We know that the total current passing through the shunt and ammeter is 5A, and the current passing through the shunt is 4.95A. Therefore, the current passing through the ammeter is: I(ammeter) = 5A - 4.95A I(ammeter) = 0.05A We also know that the resistance of the ammeter is 5Ω. Therefore, the voltage drop across the ammeter is: V(ammeter) = I(ammeter) x R(ammeter) V(ammeter) = 0.05A x 5Ω V(ammeter) = 0.25V Since the shunt resistor is connected in parallel with the ammeter, the voltage drop across the shunt resistor will also be 0.25V. Therefore, we can use Ohm's Law to find the resistance of the shunt resistor: R(shunt) = V(shunt) / I(shunt) R(shunt) = 0.25V / 4.95A R(shunt) = 0.0505Ω Therefore, the value of the resistor required to adapt the ammeter to measure a current of 5 A is 0.0505Ω, which is closest to the option 0.05Ω.