Question

A resistance of 10 Ω(R1) is connected in series with two resistances each of 15 Ω(R2 and R3) arranged in

parallel. What resistance(R4) must be shunted across this parallel combination so that the total current taken

shall be 1.5 A with 20 V applied ?​

Answer 1

Answer:

Explanation:

1/Rp = 1/R1 + 1/R2 +1/R3 +...

R1= 10

R2=15

R3=15

1/10 + 1/15 + 1/15

=> 1x15 / 10x15

1x10 / 15x10

1x10 / 15x10

= 15+10+10/150

=> 40/150

1/R = 40/150

=> R=150/40

=3.75A

Resistance seen from power supply is R1= 20/1.5=13.333 ohms

Substracting 10 ohms resistor in serie R1-10= 3.333 ohms=Rp or parallels resistance.

the two 15 ohms in parallel gaves half resistance 7.5 ohms

Rp=7.5*R / (7.5+R) or (7.5+R)*Rp=7.5*R solving for R

7.5 *Rp=R*(7.5-Rp) or R=7.5*Rp/(7.5-Rp) replacing Rp=10/3

R=75/(22.5-10)=75/12.5=6 ohms