two resistors of resistance 10 ohm and 15 ohm are connected in series to a 5V battery. calculate the current through the conductors and heat produced in 5 minutes
Answers
Explanation:
Resistor 1= R1= 10ohm
Resistor 2= R2=15ohm
Potential Difference= 12v
A) Minimum Current flowing through the circuit
when the two resistors are connected in series
Rs= R1+R2
Rs= 10+15
Rs= 25ohm
Current (I)= V/Rs
=12/25
=0.48A
B) Maximum current flows through the circuit when it is connected in parallel
Rp= R1×R2/R1+R2
Rp= 150/25
Rp= 6ohm
Current (I)=V/Rp
=12/6
=2A
Explanation:
Let R1 be 10 ohms and R2 be 15 ohms resistors,
Since R1 and R2 are connected in series
Therefore, total resistance (R) =R1 +R2= 10 +15=25ohms
Now
Current (I) = V/R = 5/25 =1/5 Amperes
Given, time (t) = 5 minutes = 300 seconds
Therefore
Heat produced =I^2Rt
= (1/5)^2 *25*300
=1/25*25*300
=300 Joules
Second method
Work done(W) = energy(E) or Heat, since heat is also a form of heat (work energy theorem)
W = V*Q(charge)
E= V*Q
= 5*It (Q=It)
= 5 * 1/5*300
=5*60
=300 joules