Consider the circuit shown below. (Assume R 1 = 1.70 , R 2 = 4.40 , R 3 = 6.20 ,

Question

Consider the circuit shown below. (Assume

R1 = 1.70 ,

R2 = 4.40 ,

R3 = 6.20 ,

and

R4 = 12.0 .)

(a) Determine the equivalent resistance of the three resistors in parallel.


(b) Determine the equivalent resistance of all four resistors in the circuit.


(c) Determine the current supplied by the battery.
A

(d) Determine the magnitude of the potential difference across the resistor R1.
V

(e) Determine the magnitude of the potential difference across the resistor R4.
V

(f) Determine the current through the resistor R4.
A

Explanation / Answer

part a )

1/R = 1/R2 + 1/R3 + 1/R4

1/R = 1/4.40 + 1/6.20 + 1/12

R = 2.12 ohm

part b )

R1 in series with R

Req = R1 + R

Req = 1.70 + 2.12 = 3.82 ohm

part c )

I = V/Req

I = 24/3.82 = 6.28 A

part d )

V = IR

current in R1 = 6.28 because in series current reamin same

V = 1.7 * 6.28 = 10.676 V

part e )

potential difference ar R ( equivalent of R2 , R3 , R4 )

V = 2.12 * 6.28 = 13.3136

they all three in parallel combination , in parallel combination voltage remain same

at R4 , V = 13.3136 V

part f )

I = V/R

I = 13.3136/12 = 1.109 = 1.11 A

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects