You are given several identical resistances each of value $R=10\, \Omega$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistances of $5\, \Omega$ which can carry a current of 4 ampere. The minimum number of resistances of the type R that will be required for this job is
- 4
- 10
- 8
- 20
The Correct Option is C
Solution and Explanation
Let r be the resistance of each path. These are connected in parallel. Hence their equivalent resistance will be r/4.
According to the given problem $\frac{r}{4}=5$ or $r = 20\, \Omega.$
For this purpose two resistances should be connected. There are four such combinations.
Hence, the total number of resistance $= 4 \times 2 = 8$
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Concepts Used:
Current Electricity
Current electricity is defined as the flow of electrons from one section of the circuit to another.
Types of Current Electricity
There are two types of current electricity as follows:
Direct Current
The current electricity whose direction remains the same is known as direct current. Direct current is defined by the constant flow of electrons from a region of high electron density to a region of low electron density. DC is used in many household appliances and applications that involve a battery.
Alternating Current
The current electricity that is bidirectional and keeps changing the direction of the charge flow is known as alternating current. The bi-directionality is caused by a sinusoidally varying current and voltage that reverses directions, creating a periodic back-and-forth motion for the current. The electrical outlets at our homes and industries are supplied with alternating current.