In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows with full intensity then what will be the equivalent resistance between X and Y

(1) 0.45 Ω

(2) 1 Ω

(3) 3 Ω

(4) 5 Ω

Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through R in the second circuit is $\frac{1}{10}$th of current through R in the first circuit. If R is 11 Ω, the value of R₁

(1) 9.9 Ω

(2) 11 Ω

(3) 8.8 Ω

(4) 7.7 Ω

In order to quadruple the resistance of a uniform wire, a part of its length was uniformly stretched till the final length of the entire wire was 1.5 times the original length, the part of the wire was fraction equal to

(1) 1 / 8

(2) 1 / 6

(3) 1 / 10

(4) 1 / 4

In the circuit shown in figure reading of voltmeter is V₁ when only S₁ is closed, reading of voltmeter is V₂ when only S₂ is closed and reading of voltmeter is V₃ when both S₁ and S₂ are closed. Then

(1) V₃ > V₂ > V₁

(2) V₂ > V₁ > V₃

(3) V₃ > V₁ > V₂

(4) V₁ > V₂ > V₃

Current through wire XY of circuit shown is :

(1) 1 A

(2) 4 A

(3) 2 A

(4) 3 A

\(12\) cells each having the same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two similar cells which are in series. Current is \(3~\text{A}\) when cells and battery aid each other and is \(2~\text{A}\) when cells and battery oppose each other. The number of cells wrongly connected is/are:
1. \(4\)
2. \(1\)
3. \(3\)
4. \(2\)

Following figure shows cross-sections through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A, and conductor C will fit snugly within conductor B. Relationship between their end to end resistance is

(1) R_A = R_B = R_C

(2) R_A > R_B > R_C

(3) R_A < R_B < R

(4) Information is not sufficient

In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be

(1) 1.944 r

(2) 0.973 r

(3) 0.486 r

(4) 0.243 r

In the adjoining circuit diagram each resistance is of 10 Ω. The current in the arm AD will be

1. $\frac{2i}{5}$

2. $\frac{3i}{5}$

3. $\frac{4i}{5}$

4. $\frac{i}{5}$

In the circuit of adjoining figure the current through 12 Ω resister will be

(1) 1 A

(2) $\frac{1}{5}A$

(3) $\frac{2}{5}A$

(4) 0 A