Question

(a) A cell of emf `3.4 V` and internal resistance `3 Omega` is connected to an ammeter having resistance `2 Omega` and to an external resistance `100 Omega`. When a voltmeter is connected across the `100 Omega` resistance the ammeter reading is `0.04 A`. Find the voltage read by the voltmeter and its resistance. Had the voltmeter been an ideal one, what would have its reading ?

(b) Two resistances of `100 Omega` and `200 Omega` are connected in series with battery of emf `4 V` and negligible internal resistance. A voltmeter of resistance `200 Omega` is used to measure voltemeter of resistances separately. Calculate the voltmeter indicated.
( c) In the circuit shown, a voltmeter reads `30 V` when it is connected across the `400 Omega` resistance. Calculate what the same voltmeter will read when it is connected across the `300 Omega` resistance ?

(d) An ammeter and a voltmeter are connected in series to a battery of emf `E = 6.0 V`. When a certain resistance is connected in parallel with the voltmeter. the reading of the ammeter is doubled. while voltmeter reading is reduced to half. Find the voltmeter reading after the connection of the resistance.

Answer 1

(a)
`3.4 = 0.04 xx R_A + 0.04 xx 3 + V_1`
=`0.04 xx 2 + 0.04 xx 3 + V_1`
=`0.2 + V_1`
`V_1 = 3.2 V`
Current in `100 Omega, (3.2)/(100) = 0.032 A`
Current through voltmeter `= 0.040 - 0.032 = 0.08 A`
Let resistance of voltmeter is `R_V`
`V_1 = 0.08 R_V`
`3.2 = 0.08 R_V`
`R_V = 40 Omega`
If voltmeter is ideal, `R_V = oo` , no current will flow through it.
`R_(eq) = 3 + 2 + 100 = 105 Omega`
Current in circuit `i = (3.4)/(105) A`
`V_X - V_Y = i xx 100 = (3.4)/(105) xx 100 = 3.23 V = V_1`
(b) (i)
`V_1 = (200//3)/(200//3 + 200) xx 4 = 1 V` : reading of voltmeter
(ii)
`V_1 = 2 V` : reading of voltmeter.
( c)
`p.d` across `300 Omega = 30 V`
Current in `300 Omega i = (30)/(300) = (1)/(10) A`
Current in `400 Omega i_1 = (30)/(400) = (3)/(40) A`
Hence current through voltmeter
`i_V = i - i_1 = (3)/(10) -(3)/(40) = (1)/(40) A`
`p.d` across voltmeter
`30 = i_V R_V = (1)/(40) R_V`
`R_V = 1200`
Resistance of voltmeter `= 1200 Omega`
Voltmeter across `300 Omega`.

`V_X - V_Y = (240)/(240 + 400) xx 60 = 22.5 V`
Reading of voltmeter `= 22.5 V`
(d)
Case (a), `V_A + V_V = 6` ....(i)
Case (b), `2 V_A + (V_V)/(2) = 6` ...(ii)
Solving `V_V = 4 V`
Reading of voltmeter in case (b)
`(V_V)/(2) = (4)/(2) = 2 V`.