You are given several identical resistances each of value R = 10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5 Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this job 

(1) 4

(2) 10

(3) 8

(4) 20

The resistance of a wire is 10^–6 Ω per metre. It is bend in the form of a circle of diameter 2 m. A wire of the same material is connected across its diameter. The total resistance across its diameter AB will be

(1) $\frac{4}{3}\pi \times {10}^{-6}\Omega$

(2) $\frac{2}{3}\pi \times {10}^{-6}\Omega$

(3) $0.88\times {10}^{-6}\Omega$

(4) $14\pi \times {10}^{-6}\Omega$

In the figure shown, the capacity of the condenser C is $2\mu F$. The current in 2 Ω resistor is : 

(1) 9 A

(2) 0.9 A

(3) $\frac{1}{9}A$

(4) $\frac{1}{0.9}A$

When the key K is pressed at time t = 0, which of the following statements about the current I in the resistor AB of the given circuit is true?

(1) I = 2 mA at all t

(2) I oscillate between 1 mA and 2mA

(3) I = 1 mA at all t

(4) At t = 0 , I = 2 mA and with time it goes to 1 mA

There are three resistance coils of equal resistance. The maximum number of resistances you can obtain by connecting them in any manner you choose, being free to use any number of the coils in any way is :

(1) 3

(2) 4

(3) 6

(4) 5

If in the circuit shown below, the internal resistance of the battery is 1.5 Ω and V_P and V_Q are the potentials at P and Q respectively, what is the potential difference between the points P and Q 

(1) Zero

(2) 4 volts (V_P > V_Q)

(3) 4 volts (V_Q > V_P)

(4) 2.5 volts (V_Q > V_P)

Two wires of resistance R₁ and R₂ have temperature coefficient of resistance ${\alpha }_1\text{ and }{\alpha }_2$, respectively. These are joined in series. The effective temperature coefficient of resistance is :

(1) $\frac{{\alpha }_1+{\alpha }_2}{2}$

(2) $\sqrt{{\alpha }_1{\alpha }_2}$

(3) $\frac{{\alpha }_1{R}_1+{\alpha }_2{R}_2}{R_1+R_2}$

(4) $\frac{\sqrt{R_1{R}_2{\alpha }_1{\alpha }_2}}{\sqrt{R_1^2+R_2^2}}$

Two cells of equal e.m.f. and of internal resistances, r_1, and $r_2(r_1>r_2)$ are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be :

(1) $r_1+r_2$

(2) $r_1-r_2$

(3) $\frac{r_1+r_2}{2}$

(4) $\frac{r_1-r_2}{2}$

When connected across the terminals of a cell, a voltmeter measures 5V and a connected ammeter measures 10 A of current. A resistance of 2 ohms is connected across the terminals of the cell. The current flowing through this resistance will be :

(1) 2.5 A

(2) 2.0 A

(3) 5.0 A

(4) 7.5 A

In the circuit shown here, E₁ = E₂ = E₃ = 2 V and R₁ = R₂ = 4 ohms. The current flowing between points A and B through battery E₂ is 

(1) Zero

(2) 2 amp from A to B

(3) 2 amp from B to A

(4) None of the above