XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Chapter # 15 PROF. IMRAN HASHMI PROBLEMS FROM TEXT BOOK Q.1 A galvanometer has a resistance of 50 ohms and it deflects full scale when a current of 10 milliamperes flows in it. How can it be converted into an ammeter of range 10A? Solution: Rg = 50 ο Rg = Ig = 10 mA = 10 x 10-3 A = 0.01A [A current of Ig = 10 mA is to be used to I = 10A measure a current of I = 10 A using a shunt Rs = ? resistance of Shunt resistance Rs = 0.05 ο. The excess current of 9.99A (Is = I π°π πΉπ Rs = π°−π°π = π.πππ π ππ ππ−π.ππ 50 Ω – Ig) can detour around the 50-ohm (= Rg) galvanometer coil]. [Vide Fig. 15.16 (a)] Rg = π.π π.ππ Q.2 A galvanometer whose resistance if 40 ohms deflects full-scale for a potential difference of 100 millivolts across its terminals. How can it be converted into an ammeter of 5 ampere range? Solution: Now Rg = 40 ο Ig = Vg = 100mV = 100 x 10-3 = 0.1 V π½π πΉπ Rs = I = 5A = π.π ππ π°π πΉ π π°−π°π = 2.5 x 10-3A = 0.0025 A = π.ππππ π ππ π−π.ππππ = π.π π.ππππ Rs = ? Rs = 0.02ο Q.3 The coil of a galvanometer which has a resistance of 50 ohms and a current of 50 microamperes produces full scale deflection in it. Show by a diagram how can it be converted to (a) an ammeter of 5 ampere range and compute the shunt resistance (b) a voltmeter of 300 volts range and compute the series resistance. (Multiplier resistance). Solution: Rg = 50ο (b) The voltage V across the Ig = 500 µA = 500 x 10-6 A series combination of I = 5A galvanometer coil and a Rs = ? multiplier resistor is V = 300 V I(Rg + Rx). So the multiplier Rx = ? resistance, Rx (a) Shunt resistance Rs = Rs = ππ π π Rx = π−ππ πππ π ππ−π π ππ π−(πππ π ππ−π ) Rs = 0.005ο = π = ( ) − πΉπ πππ (πππ π ππ−π ) ππ - (50) = 600000 - 50 π.πππ π.ππππ Rs = 599950 ο XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Q.4 Chapter # 15 PROF. IMRAN HASHMI A galvanometer of resistance 25 ohms deflects full scale for a current of 0.05 ampere. It is desired to convert this galvanometer into an ammeter reading 25 amperes full scale. The only shunt available is 0.06 ohm. What resistance R must be included in series with the galvanometer coil, as shown in figure, for using this shunt? Solution: Rg = 25 ο Ig = 0.05 A I = 25 A Rs = 0.06 ο R=? Vab = Ig R + Ig Rg = Ig (R + Rg) ……………….(i) Vab = (I - Ig) Rs ………………….(ii) comparing eq. (1) & (2) Ig (R + Rg) = (I - Ig) Rs R + Rg = ( π°−π°π π°π ) πΉπ πΉ + πΉπ = ( ππ − π. ππ π. πππ ) π π. ππ = π. ππ π. ππ R + Rg = 29.94 R = 29.94 – Rg R = 29.94 – 25 R = 4.94 Ω Q.5 An ammeter deflects full scale with a current of 5 amperes and has a total resistance of 0.5 ohms What shunt resistance must be connected to it to measure 25 amperes full scale? Solution: The initial ‘ammeter’ is treated as a galvanometer. Ig = 5A Rg = 0.5 ο I = 25 A Rs = RS = RS = π°π πΉπ π°−π°π π π π.π ππ−π ππ ππ Rs = ? RS =0.125ο XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Q.6 Chapter # 15 A moving coil galvanometer has a resistance of 50 ohms and deflects full scale with a current of 0.005 amperes. What resistance R1, R2 and R3 must be connected to it as shown in figure to measure currents up to 1A, 5A and 10A? Solution: Rg = 50ο Ig = 0.005 A (i) For 10A range R2 and R3 in series with G; and R1 will be the shunt. Now I = 10A. [vide Fig. 15.19] Formula: shunt, R1 = R1 = PROF. IMRAN HASHMI π°π (πΉπ +πΉπ +πΉπ ) (iii) For 1A range R1 + R2 + R3 will be the shunt. [vide Fig. 15.18] π°π πΉπ ο R1 + R2 + R3 = π°− π° π π.πππ π ππ R1 + R2 + R3 = (π−π.πππ) π.πππ R1 + R2 + R3 = π.πππ R1 + R2 + R3 = 0.251 (3) Now, 1999 R1 – R2 – R3 = 50 R1 + R2 + R3 = 0.251 (3) (π°−π°π ) [By adding eq. (1) and eq. (3)] 2000 R1 = 50.251 π.πππ(ππ+πΉπ+πΉπ) (ππ−π.πππ) 9.995 R1 = 0.005 (50 + R2 + R3) ο 1939 R1 – R2 – R3 = 50 π.πππ πΉπ π.πππ = R2 + R3 + 50 R1 = (1) ππ.πππ ππππ = Also, 999 • (1) R1 /\/\/\/ R2 • /\/\/\/\ Rg G R3 • /\/\/\/\ • 0.0251 ο R1 + 999 R2 – R3 = 50 R1 + R2 + R3 = 0.251 (2) (3) [By adding] 1000 R1 + 1000 R2 = 50.251 (1000 x 0.0251) + 1000 R2 = 50.251 ππ.πππ 1000 R2 = 50.251 – 25.1 ο R2 = ππππ ο R2 = 0.0251 ο + 10A 5A 1A (ii) For 5A range R3 in series with G; and “R1 + R2” will be the shunt. [vide Fig. 15.20] R1 + R2 = R1 + R2 = R3 = 0.2008 ο G π°π (πΉπ +πΉπ ) (π°−π°π ) π.πππ(ππ+πΉπ) /\/\/\/\/ R1 (π−π.πππ) 4.995 (R1 + R2) = 0.005 (50 + R3) π.πππ (R1 + R2) = 50 + R3 ο 999R1 + 999R2 – Ra = 50 (2) π.πππ R2 R3 /\/\/\/ \/\/\/\ G /\/\/\/\/\/\/\/\/ R1 + Putting values of R1 and R2 in eq. (3), we get 0.0251 + 0.0251 + R3 = 0.251 ο R3 = 0.251 – 0.0502 10A R3 /\/\/\/\/\/ /\/\/\/\/\/\/\ R2 5A + XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Q.7 Chapter # 15 PROF. IMRAN HASHMI A 300-volt voltmeter has a total resistance of 20,000 ohms. What additional series resistance must be connected to it to increase its range to 500 volts? Solution: Rx = The initial voltmeter is regarded as a π½π π°π πππ − πΉπ (π.πππ) - 20,000 = 33333.3 – 20,000 Rx = 13333.3 ο galvanometer. Vg = 300 volt, Rg = 20,000ο, Rx = ? V = 500v π½π πππ Now Ig = πΉ = ππ,πππ = 0.015 A π Q.8 The resistance of a moving-coil galvanometer is 25 ohms and a current of 1 milliampere causes full scale deflection in it. It is to be converted into a multi-range voltmeter. Find the series resistances R1, R2 and R3 to give the ranges of 5V, 50V and 500V at the range terminals, as shown in the figure. Solution: Data: Rg = 25 ο Ig = 1 x 10-3 A. (i) For 5V range Multiplier, Rx = R1 π½ π ο πΉπ = π° − πΉπ = (π π ππ−π ) − ππ (ii) For 50V range Rg = Rg + R1 Multiplier = R2 π½ ππ πΉπ = ( ) − (πΉο’ο’π ) = ( ) − (πΉπ + πΉπ ) π°π π π ππ−π R2 = 50000 – (25 + 4975) R2 = 50,000 – 5000 = 45,000 ο π (iii) For 500V range Rgο’ο’ = Rg + R1 + R2 Multiplier =R3 R1 = 4975 ο π½ πππ πΉπ = (π° ) − πΉο’ο’π = π π ππ−π − (πΉπ + πΉπ + πΉπ ) π G R1 /\/\/\/ R2 \/\/\/\ R3 = 500,000 – (25 + 4975 + 45000) = R3 /\/\/\/\ + 5v 50v 500v 450,000 ο XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Chapter # 15 PROF. IMRAN HASHMI The galvanometer of the ohmmeter in the figure has a resistance of 25Ω and deflects full scale with a current of 2 milli-amperes in it. The e.m.f. of the cell is 1.5 volts. Q.9 (i) What is the value of the series resistor R? (ii) To what value of X connected to its terminal do the deflection of 1/5, full-scale correspond? (iii) Is the scale of the ohmmeter linear? Solution: (i) Rg = 25Ω E = 1.5V Ig = 2 x 10-3A (iii) The scale of the ohmmeter is not linear, because its resistance is not proportional to the deflection. R G • /\/\/\/\/\ π½ π¬ π π G π.π R = (π° ) − πΉπ = (π° ) − πΉπ = (π π ππ−π ) − ππ < x < < R = 0.75 x 103- 25 = 750 – 25 R = 725 ο π X1 = π½ π°π − (πΉ + πΉπ ) = ( π.π π.π π ππ + • π (ii) For πth deflection, Ig = π x 2 x 10-3 = 0.4 x 10-3A E ) ) ( −π − πππ π ππ X1 = ππππ − πππ X1 = 3000 ο Q.10 A potentiometer is set up to measure the emf Ex of a cell. The potentiometer wire is 120cm long. Es is the e.m.f. of a standard cadmium cell, equal to 1.018V. When the key 1 only is closed to include the emf Ex in the galvanometer circuit, the galvanometer gives no deflection with the sliding contact at C, 56.4cm from A. When the key 2 only is closed to include the emf Es in the galvanometer circuit, the balance is obtained at C, 43.2 cm from A. (a) What is the emf Ex of the cell? (b) What is the p.d. across the entire length of the wire AB? Solution: π π ππ ππ.π (a) π π± = ππ ο π.πππ = ππ.π π π Ex = 1.108 x 1.3055 = 1.329 V π π π πππ (b) π = π ο π.πππ = ππ.π π¬ π¬ V = 2.828volts • A C Cο’ G Ex Es Q.11 1 ( ) 2 ( ) • B • A certain galvanometer has a resistance of 40 ohm and deflects full scale for a voltage of 100mV across its terminal. How can it be made into 2-A ammeter? π½π πππ π ππ−π Solution: Ig = πΉ = ππ π Rg = 40Ω -3 I = 2.5 x 10 A = 0.0025 A g Vg = 100mV (for f.s.d.) π π π.π π± ππ−π π± ππ π.π π π I = 2A Rs = (π− π ) = (π−π.ππππ) = π.ππππ = 0.05Ω π Rs = ? Now Current through galvanometer for f.s.d. XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Chapter # 15 PROF. IMRAN HASHMI Q.12 A certain galvanometer has a full scale deflection for 0.003A. How can this glavanometer of 20 ohm resistance be converted into 90-V voltmeter? π½ ππ Solution: ) − (ππ) Rx = ( ) − πΉπ = ( π°π π.πππ Data: Rg = 30000 - 20 Ig = 0.003A Rg = 20Ω Rg = 29980 Ω V = 90V Rx = ? Q.13 If the resistance in three successive arms of a balanced bridge be 1, 2 and 36 ohms, respectively, calculate the resistance of the galvanometer in the fourth arm. πΉπ πΉπ π ππ = ο = πΉπ πΉπ π πΉπ 2 R4 = 36 x1 = 72 Ω Solution: R1 = 1Ω R2 = 2Ω R3 = 36Ω R4 = ? Q.14 A resistance of 10 ohms is placed in the right gap and one of unknown value is placed in the left gap of metre bridge. The null point is found to be at 40cm. Find the unknown resistance . Solution: πΏ ππ πΏ ππ = ο = πΉ ππ ππ ππ X = 10 x 0.666 = 6.66 Ω Q.15 The resistance of a galvanometer is 20 ohms and it gives full scale deflection with a current of 0.001 ampere. (a) What should be the resistance of the parallel resistor ised to convert it into an ammeter of 10 amperes. (b) What should be the resistance of the series resistor to convert it into a voltmeter of 10 volts. (Karachi Board, 1980) π°π πΉπ π.πππ π ππ π.ππ Solution: (a) Rs = π°−π° = (ππ−π.πππ) = π.πππ = 0.002 Ω π Rg = 20Ω Ig = 0.001A π½ ππ I = 10A (b) Rx = π° − πΉπ = (π.πππ) – 20 9980 Ω π Rs = ? V = 10V Rx = ? Q.16 A galvanometer has a resistance of 50 ohms, and gives full scale deflection for a p.d. of 150 mV. If the galvanometer is to be converted into an ammeter reading upto 4A, what should be the resistance of the parallel resistor used? Solution: π°π πΉπ Rg = 50Ω Rs = π°−π° -3 π Vg = 150mV = 150 x 10 V π.πππ π ππ I = 4A Rs = π−π.πππ Rs = ? π.ππ Rs = π.πππ Since π½π πππ π ππ−π = 3 x 10-3 = 0.003A Rs = 0.037 Ω Q.17 A galvanometer of resistance 20.0 ohms gives full scale deflection with a current of 0.005 amperes. How will you convert it to an instrument measuring a maximum current of 1.0 ampere? What will be the resistance of the instrument thus converted? (Karachi Board, 1989) Solution: Since Rg and Rs are in parallel, hence total resistance of the Rg = 20Ω galvanometer is πΉπ π πΉπ π π π Ig = 0.005 A = πΉ +πΉ ο πΉ = πΉ +πΉ I = 1.0A πΉ Ig = πΉ = ππ π π Rs = ? R=? Resistance of shunt, Rs, = π°π πΉπ π°−π°π = π.πππ π ππ π−π.πππ Rs = 0.1005 Ω R= = π.π π.πππ π ππ π π.ππππ ππ+π.ππππ π π.ππ π = ππ.ππππ = 0.1 Ω XII - PHYSICS MAGNETISM AND ELECTROMAGNETISM Chapter # 15 PROF. IMRAN HASHMI A galvanometer of resistance 60Ω gives full scale deflection with a current of 4mA. A resistance of 1094Ω is connected in series with the coil to convert it into a voltmeter. Find the range of the voltmeter obtained. (Karachi Board, 1995) Solution: ο V = (Rx + Rg) Ig Rg = 60Ω V = (10940 + 60) x 4 x 10-3 -3 Ig = 4mA = 4 x 10 A V = 11000 x 4 x 10-3 Rx = 1094Ω V = 44000 x 10-3 = 44V V=? Required range = 0V to 44V π½ ο Rx = (π° ) – Rg Q.18 π
