An unknown resistance R₁ is connected in series with a resistance of 10 Ω. This combinations is connected to one gap of a metre bridge while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. Now, when the 10 Ω resistance is removed the balance point shifts to 40 cm. The value of R₁ is (in ohm) 

(1) 60

(2) 40

(3) 20

(4) 10

A wire has a resistance of 6 Ω. It is cut into two parts and both half values are connected in parallel. The new resistance is :

(1) 12 Ω

(2) 1.5 Ω

(3) 3 Ω

(4) 6 Ω

Six equal resistances are connected between points P, Q and R as shown in the figure. Then the net resistance will be maximum between

(1) P and Q

(2) Q and R

(3) P and R

(4) Any two points

The total current supplied to the circuit by the battery is:

1. \(1~\text{A}\)
2. \(2~\text{A}\)
3. \(4~\text{A}\)
4. \(6~\text{A}\)

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be 

(1) 3

(2) 1/3

(3) 8/9

(4) 2

In circuit shown below, the resistances are given in ohms and the battery is assumed ideal with emf equal to \(3\) volt. The voltage across the resistance \(R_4\) is:

1. \(0.4\) V
2. \(0.6\) V
3. \(1.2\) V
4. \(1.5\) V

If you are provided three resistances 2 Ω, 3 Ω and 6 Ω. How will you connect them so as to obtain the equivalent resistance of 4 Ω 

(1)

(2)

(3)

(4) None of these

The equivalent resistance and potential difference between A and B for the circuit is respectively 

(1) 4 Ω, 8 V

(2) 8 Ω, 4 V

(3) 2 Ω, 2 V

(4) 16 Ω, 8 V

Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be 

(1) $\frac{3V}{R}$

(2) $\frac{V}{R}$

(3) $\frac{V}{2R}$

(4) $\frac{2V}{R}$

For the network shown in the figure the value of the current i is 

(1) $\frac{9V}{35}$

(2) $\frac{5V}{18}$

(3) $\frac{5V}{9}$

(4) $\frac{18V}{5}$