In the circuit shown in the diagram ( Figure 1 ) , suppose R 1 = R 2 = 240 ? and
ID: 2256706 • Letter: I
Question
In the circuit shown in the diagram(Figure 1) , suppose R1=R2 = 240? and R3=R. The emf of the battery is 12.0V .
Part A:
Find the value of R such that the current supplied by the battery is 0.0770A .
R=__________ohms
Part B:
Find the value of R that gives a potential difference of 3.75V across resistor 2.
R=__________ohms
In the circuit shown in the diagram(Figure 1) , suppose R1=R2 = 240? and R3=R. The emf of the battery is 12.0V . Find the value of R such that the current supplied by the battery is 0.0770A . Find the value of R that gives a potential difference of 3.75V across resistor 2.Explanation / Answer
a) The equivalent resistance of the circuit will be:
R_eq = [{1/(R2+R3)} + (1/R1)]^(-1)
= [{1/(240+R)} + (1/240)]^(-1)
= [240(240 + R)] / (480 + R)
It is given that: I = 0.0770 A
Since, V = IR
=> 12 / 0.077 = R_eq
=> 12 * (480 + R) = [240(240 + R)] * 0.077
=> 5760 - 4435.2 = (18.48 - 12) R
=> 6.48 R = 1324.8
=> R = 1324.8 / 6.48 = 204.44 ohm
b) R2 = 240 ohm, V2 = 3.75 V
=> I2 = V2 / R2 = 3.75 / 240 = 0.015625 A
Thus, the current in resistor will also be 0.015625 A (because R2 and R are in series)
Also, the net potential difference between R2 and R is 12 V
=> V3 + V2 = 12
=> V3 = 12 - 3.75 = 8.25 V
Hence, R = V3 / I3 = 8.25 / 0.015625 = 528 ohm