Why mcqs of class 12 Physics lesson 3 Current Electricity is important ?
- these mcqs are important because these questions cover a wide topic of the chapter
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- mcqs break the big topic in some small topic so they are easy to learn
- they give a short revision at the time of class 12 Physics exam
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CLASS 12 PHYSICS LESSON 3 CURRENT ELECTRICITY OBJECTIVE QUESTIONS
1. Two cells of 1.25 V and 0.75 V are connected in series with anode of one connected to cathode of the other. The effective voltage will be
2.0 V
0.75 V
1.25 V
0.50 V
Answer - A
Explanation - The cells support each other.
Eeq=E1+E2=1.25+0.75=2.0V
2. Two cells of 1.25 V and 0.75 V are connected in series with anode of one connected to anode of the other. The effective voltage will be
2.0 V
1.25 V
0.50 V
0.75 V
Answer - C
Explanation - The cells are in opposition.
Eeq=E1−E2=1.25−0.75=0.5V
3. A current of 2 A flows in conductors as shown. The potential difference VA- VB will be
+2 V
-1 V
+4 V
+1 V
Answer - D
4. The instrument for the accurate measurement of the e.m.f of a cell is
a potentiometer
a slide wire bridge
a voltmeter
an ammeter
Answer - A
Explanation - Both potentiometer and voltmeter are devices to measure potential difference. Emf is the terminal p.d between the electrodes of a cell in open circuit, i.e., when no current is drawn from it. Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.
5. Potentiometer measures the potential difference more accurately than a voltmeter, because
It has a wire of high resistance.
It has a wire of low resistance.
It draws a heavy current from an external circuit.
It does not draw current from an external circuit.
Answer - D
Explanation - Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.
6. In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by
putting a suitable resistance in series with the cell
putting a shunt resistance in parallel with the cell
increasing the current due to auxiliary battery
decreasing the current due to auxiliary battery
Answer - D
Explanation - If the current due to the auxiliary battery is decreased, the potential gradient will be decreased, so the balancing length increases.
Thus, null point will move to the fifth wire.
7. It is observed in a potentiometer experiment that no current passes through the galvanometer, when the terminals of the cell are connected across a certain length of the potentiometer wire. On shunting the cell by a 2 Ω resistance, the balancing length is reduced to half. The internal resistance of the cell is
18 Ω
4 Ω
2Ω
9 Ω
Answer - C
8. The sensitivity of the potentiometer can be increased by:
increasing the length of potentiometer wire.
decreasing the length of potentiometer wire.
increasing the e.m.f. of primary cell.
increasing the potential gradient.
Answer - A
Explanation - A potentiometer is considered to be sensitive if the potential gradient dVdl is low. Such a potentiometer can measure very small changes in potential difference. Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity, increasing the emf of the primary cell and by decreasing the length, potential gradient increases.
9. Two resistances are connected in the two gaps of a meter bridge. The balance points is 20 cm from the zero end. When a resistance of 15 ohm is connected in series with the smaller of the two resistances, the null point shifts to 40 cm. The smaller of the two resistances has the value (in ohm)
9
8
12
10
Answer - A
Explanation - Let the resistances be P and Q. When balanced,
PQ=2080=14....(i)
When another resistance is connected in series,
P+15Q=4060=23....(ii)
From (i) and (ii),
P+154P=23
⇒ 3P + 45 = 8P
⇒ 5P = 45
⇒P=9Ω
10.Kirchhoff’s second law is a consequence of law of conservation of
momentum
charge
angular momentum
Energy
Answer - D
Explanation - Work done in moving a charge in a closed circuit is zero, therefore the algebraic sum of potential differences in a closed circuit is zero.
11.In a meter bridge, when galvanometer and cell positions are interchanged,
Exactly same relation between four resistance is obtained
No relation is obtained between four resistances
Inverse relation between four resistance is obtained
Can’t predict the relation
Answer - A
Explanation - The battery in the circuit (1) shown is connected between A and C, while the galvanometer is connected between B and D. In the bridge balanced condition, PR=QS
When the battery and the galvanometer are interchanged, the circuit takes the form (2). The circuit (2) can be redrawn as circuit (3). The ratio of the resistances in the bridge balanced condition is PQ=RS which is same as, PR=QS
12. An ammeter together with an unknown resistance in series is connected across two identical batteries each of emf 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1A and when the batteries are in parallel, the current is 0.6A. Then the internal resistance of the battery is
1/4Ω
1/3Ω
1/2Ω
1/5Ω
Answer - B
Explanation - When the batteries are in series, the internal resistances are in series. Both emf and the internal resistances add up.
I=E(r+R)1=3(2r+R)⇒R+2r=3....(i)
and in parallel combination, 0.6=2×1.5(r2+R)⇒5=2R+r....(ii)
From equation (i) and (ii),
r=1/3Ω
13. A battery is connected with a potentiometer wire. The internal resistance of the battery is negligible. If the length of the potentiometer wire of the same material and radius is doubled then
Potential gradient does not change
Potential gradient becomes two times
Potential gradient becomes half
None of these
Answer - A
Explanation - Potential gradient is given by V/l. If V and l are constant, the potential gradient also remains constant. The change in the radius will cause a change in the current. This does not change the potential gradient.
14. A potential difference of 220 V is maintained across 12000 Ω rheostat. Then voltmeter V has a resistance of 6000 Ω and point C is at one fourth the distance from a to b. Then the reading of voltmeter is
45 volt
40 volt
30 volt
60 volt
Answer - B
Explanation - Resistance R = 14R = 120004 = 3000 parallel
1Rp=1Rac+1RV =13000+16000=12000; Rp=2000Ω
This is in series with Rbc.
Rbc = R - Rac = 12000 - 3000Ω = 9000Ω
Req = Rp + Rbc = 9000 + 2000 = 11000Ω.
The current in the circuit I=22011000=0.02A
The reading of the voltmeter is the potential drop across Rp, V = 0.02 × 2000 = 40 V
15.Figure shows current in a part of an electrical circuit. Then current I is
2.5 A
1.5 A
2.1 A
0.5 A
Answer - A
Explanation - Using Kirchhoff’s junction rule at junction A,
2 + 5 - 2 - 1 + IAB = 0 ; IAB = -4 Amp
The current IAB is directed away from the junction A.
Using the junction rule at B,
4 + 0.2 - 1.7 - I = 0
I = 2.5 Amp
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