1
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For the circuit shown, with R1 = 1.0W, R2 = 2.0 W, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V): JEE Main 2019 (Online) 8th April Morning Slot Physics - Current Electricity Question 214 English
A
3.3
B
2.7
C
2.3
D
3.7
2
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 $$ \times $$ 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :
A
200 ohm
B
250 ohm
C
6200 ohm
D
6250 ohm
3
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the given circuit diagram, the currents, I1 = – 0.3 A, I4 = 0.8 A and I5 = 0.4 A, are flowing as shown. The currents I2, I3 and I6, respectively, are :

JEE Main 2019 (Online) 12th January Evening Slot Physics - Current Electricity Question 216 English
A
1.1 A, – 0.4 A, 0.4 A
B
$$-$$ 0.4 A, 0.4 A, 1.1 A
C
0.4 A, 1.1 A, 0.4 A
D
1.1 A, 0.4 A, 0.4 A
4
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The galvanometer deflection, when key K1 is closed but K2 is open, equals $$\theta $$0 (see figure). On closing K2 also and adjusting R2 to 5$$\Omega $$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Current Electricity Question 221 English
A
5 $$\Omega $$
B
25 $$\Omega $$
C
12 $$\Omega $$
D
22 $$\Omega $$
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