Series and parallel combinations of resistors — NEET & JEE Questions

Question 1 · difficulty L1 · recall

For resistors connected in series, the equivalent resistance is

• A. the sum of the individual resistances
• B. less than the smallest resistance
• C. the reciprocal of the sum
• D. always zero

Question 2 · difficulty L2 · understanding

A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Question 3 · difficulty L2 · understanding

A wire of resistance $$20 \Omega$$ is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ $$\Omega$$.

Question 4 · difficulty L2 · understanding

Equivalent resistance of the following network is __________ $$\Omega$$.

Question 5 · difficulty L2 · understanding

10 resistors each of resistance 10 $$\Omega$$ can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ___________.

Question 6 · difficulty L2 · understanding

The current I flowing through the given circuit will be __________A.

Question 7 · difficulty L2 · understanding

A square shaped wire with resistance of each side 3$$\Omega$$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $$\Omega$$ will be ___________.

Question 8 · difficulty L2 · understanding

In the given figure switches S 1 and S 2 are in open condition. The resistance across ab when the switches S 1 and S 2 are closed is _____________ $$\Omega$$.

Question 9 · difficulty L2 · understanding

The voltage across the 10$$\Omega$$ resistor in the given circuit is x volt. The value of 'x' to the nearest integer is _________.

Question 10 · difficulty L2 · understanding

There are ' $n$ ' number of identical electric bulbs, each is designed to draw a power $p$ independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is :

• A. np
• B. p
• C. $\frac{\mathrm{p}}{\mathrm{n}^2}$
• D. $\frac{p}{n}$