Question

Two resistance R1=30 Ω and R2=20 Ω are connected in series to a 10V battery. What is the heat dissipated in R2 during a1/2 h?

Answer 1

Heat dissipated, H = 1440 Joules

Step-by-step explanation:

It is given that,

Resistor 1,
`R_1=30\ ohms`

Resistor 2,
`R_2=20\ ohms`

Voltage, V = 10 volts

Both resistors are connected in series. The equivalent resistance is given by :

`R=R_1+R_2`

R = 30 + 20

R = 50 ohms

In series combination, the current throughout the circuit remains the same. It can be calculated using ohm's law as :

V = IR

`I=(V)/(R)`

`I=(10)/(50)`

I = 0.2 A

Heat dissipated in R2 during (1/2) h, t = 1800 s is given by :

`H=I^2R_2 t`

`H=(0. 2)^2* 20* 1800`

H = 1440 J

So, the heat dissipated during 1800 seconds is 1440 Joules. Hence, this is the required solution.