A 24 ohm resistor connected across a 6.0 V battery of negligible internal resist

Question

A 24 ohm resistor connected across a 6.0 V battery of negligible internal resistance. Calculate the following: Show calculation: The total current flowing in the circuit: ______ The total power consumed by the two resistors: _______ Two resistors R_1 (12 ohm) and R_1 (24 ohm) are connected in series across a 6.0 V battery of negligible internal resistance. Draw a circuit diagram (to the right) and calculate: The total resistance of the two resistors: ______ The total current flowing in the circuit: _____ The current flowing in R_1 ______ The current flowing in R_2 ______ The total power consumed by R_1 and R_2 ______ Two resistors R_1 (12 ohm) and R_1 (24 ohm) are connected in parallel across a 6.0 V battery of negligible internal resistance. Draw a circuit diagram (to the right) and calculate: The total resistance of the two resistors _____ The total current flowing in the circuit _____ The current flowing in R_1 ______ The current flowing in R_2 _______ The

Explanation / Answer

1 )

using Ohm's Law V = i R

i = V / R

i = 6 / 24 = 0.25 A

to calculate the power

P = V X i

P = 6 X 0.25

P = 1.5 W

2 )

here the resistors are connected in series

R = R1 + R2

R = 12 + 24

R = 36 ohm

the current is V = i R

i = V / R

i = 6 / 36

i = 0.167 A

current acorss the 12 ohms

is same as the current cross the 24 ohms

that is i = 0.167 A

the Power is P = i2 R

P = 0.1672 X 36

P = 1.0040 W

3 )

when the resistors are connected in parallel

1 / R = 1 / R1 + 1 / R2

R = 24 / 3

R = 8 ohm

using ohm's law

V = i R

i = V / R

i = 6 / 8

i = 0.75 A

current through 12 ohm is

i1 = V / R1

i1 = 6 / 12

i1 = 0.5 A

and

i2 = V / R2

i2 = 6 / 24

i2 = 0.25 A

4 )

i = 1.5 A , V = 120 V

P = V X i

P = 1.5 X 120

P = 180 W

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.