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Question: If the current is flowing through a 10 ohm resistor, then indicate in which case the maximum heat wi...

If the current is flowing through a 10 ohm resistor, then indicate in which case the maximum heat will be generated?
(A) 5 ampere in 2 minutes
(B) 4 ampere in 3 minutes
(C) 3 ampere in 6 minutes
(D) 2 ampere in 5 minutes

Explanation

Solution

Heat generated is related to the energy consumed by the resistor. The energy consumed is directly proportional to the square of the electric current. Check which of the options will consume the most energy.
Formula used: In this solution we will be using the following formulae;
H=I2RtH = {I^2}Rt where HH is the heat generated by the resistor, II is the amount of current flowing through the resistor, RR is the resistance of the resistor and tt is the time for which the current flowed through the resistor.

Complete Step-by-Step solution:
The heat generated is simply the electrical energy consumed by the resistor.
Hence, to solve the above question, we calculate the energy consumed in each case of the options one after the other then check for which of them is maximum.
Hence the heat generated can be given as
H=I2RtH = {I^2}Rt where HH is the heat generated by the resistor, II is the amount of current flowing through the resistor, RR is the resistance of the resistor and tt is the time for which the current flowed through the resistor.
Option A, current 5 ampere in 2 minutes
Ha=52(10)2=500J{H_a} = {5^2}\left( {10} \right)2 = 500J
For option B,
Hb=42(10)3=480J{H_b} = {4^2}\left( {10} \right)3 = 480J
For option C,
Hc=32(10)6=540J{H_c} = {3^2}\left( {10} \right)6 = 540J
And for option D,
Hd=22(10)5=200J{H_d} = {2^2}\left( {10} \right)5 = 200J
As can be observed, the maximum heat generated is option C.

Hence, the correct option is option C

Note: Alternatively, since the resistor is the same in all cases, we can simply find the heat generated per unit ohms in each case. As in
For option A,
ha=52×2=50J/Ω{h_a} = {5^2} \times 2 = 50J/\Omega where hh is the heat generated per unit resistance.
For option B,
hb=42×3=48J/Ω{h_b} = {4^2} \times 3 = 48J/\Omega
For option C,
hd=32×6=54J/Ω{h_d} = {3^2} \times 6 = 54J/\Omega
And for option D
hd=22×5=20J/Ω{h_d} = {2^2} \times 5 = 20J/\Omega
And this still shows that option C is the correct option.