A student uses the resistance of a known resistor ( 1 Ω ) to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be 1 × 10 3 Ω in circuit (a) and 0 . 999 Ω in circuit (b). From these measurements, the resistance (in Ω ) of the voltmeter and ammeter are found to be close to

Step-by-step solution

When a voltmeter put in series, it still reads potential drop and when an ammeter is connected in parallel, it still shows current through it. Case a Let I = current through cell, then potential drop read by voltmeter is V = I · R V (this is reading of voltmeter) Where, R V is the resistance of voltmeter In loop A B , V A B = I 1 × 1 = I 2 × R A and I = I 1 + I 2 Where, R A is the resistance of ammeter We substitute for I 1 from above equation to get ⇒ I = I 2 R A + I 2 = I 2 R A + 1 ⇒ I 2 = I R A + 1 (this is read