Chapter 10, Problem 4TC

Rank the current in three branches and the current in battery, from greatest to least.

Solution

To determine:Rank the current in three branches and the current in battery, from greatest to least.

Answer:

The amount of current from greatest to least order is battery, bulb C, bulb B and bulb A.

Explanation:

Given info: Resistance of bulb A is $3\Omega$ , B is $2\Omega$ and for C is $1\Omega$ .

Three bulbs A, B and C connected in series with a battery, assuming the voltage of battery D is V.

Write the expression for the current.

$I=\frac{V}{R}$

Here,

V is voltage

R is resistance

Current and resistance are varies inversely to each other and voltage is same. For the maximum value of resistance, the current will be minimum and for the minimum value of resistance, current will be maximum.

Arranging the bulbs according to descending order of resistance is A, B and C. Thus, the current flow from them is in ascending order as A, B and C. The resistance of the battery is minimum from all bulbs, so the current in battery is also maximum.

Conclusion:

Therefore, the amount of current from greatest to least order is battery, bulb C, bulb B and bulb A.