Question

A moving coil instrument gives full scale deflection for 1 mA and has a resistance
of 5 ohms. If a resistance of 0.55 ohms is connected in parallel to the instrument,
what is the maximum value of current it can measure?​

Answer 1

Answer: I think so the answer must be approx. 0.01A

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Explanation: We know,

                      V = IR

                    ⇒ V = 0.001 × 5 = 0.005V  ------  I

         Now,    Equi. R = (0.55 × 5) ÷ (0.55 + 5)

                                  = 55 / 111 Ω

            ∴   V = IR

            ⇒ 0.005 = I × 55/111

            ⇒ I ≈ 0.01 A

Answer 2

Answer:

The maximum value of current it can measure = 10 mA

Explanation:

• Step-1: Initial resistance, R1 = 5 ohm

                    New added resistance , R2 = 0.55 ohm

     After adding the resistance in parallel, the equivalent resistance = R

     So, $\frac{1}{R}$ = $\frac{1}{R1}$ + $\frac{1}{R2}$

               = $\frac{1}{5}$ + $\frac{1}{0.55}$

                 = $\frac{111}{55}$                      So, R = 55/111

• Step-2: When Current = I = 1 mA = 1/1000 A

                    Then Resistance = R1 = 5 ohm

                     Potential difference, V = IR = $\frac{1}{1000}$ × 5 = 1/200 V

• Step-3: Voltage will remain same, V= 1/200 V

                     Resistance = R = 55/111

                     Current, I = V/R = $\frac{1}{200}$ ÷ $\frac{55}{111}$ = 0.0101 A = 10 mA

         ∴The maximum value of current it can measure = 10 mA