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?

Heat dissipated, H = 1440 Joules

Explanation:

It is given that,

Resistor 1,

Resistor 2,

Voltage, V = 10 volts

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

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 = 0.2 A

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

H = 1440 J

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