1) A 30 V battery is connected to two resistors in series. One resistor is 5 ohms, and the other resistor is 10 ohms.
a.) What is the total resistance for the circuit?
b.) What is the current passing through each resistor?
c.) What is the power used (in kW) for the 5-ohm resistor?
d.) What is the power used (in kW) for the 10-ohm resistor?
e.) If the battery runs for 20 hours, and its costs $2.50/kW*hr, what will the total charge be?
2) A 30 V battery is connected to two resistors in parallel. One resistor is 5 ohms, and the other resistor is 10 ohms.
a.) What is the total resistance for the circuit?
b.) What is the current passing through each resistor? It is different for each resistor.
c.) What is the voltage passing through each resistor? It is the same for each resistor.
c.) What is the power used (in kW) for the 5-ohm resistor?
d.) What is the power used (in kW) for the 10-ohm resistor?
e.) If the battery runs for 20 hours, and its costs $2.50/kW*hr, what will the total charge be?
Answers
Answer:
The total resistance of the circuit
=8+(20*30/50)
=8+600/50
=8+12
=20 Ohms
The total current of the circuit
I=12/20
I=0.6 Amp.
The total current flow through the 8-ohm resistor and divided into two parts in 20 and 30 Ohms resistance. Let the current flowing in the 20 Ohms resistance is I1.
I1=0.6*30/(20+30)
=0.6*30/50
=0.36 Amp
Given : A 30 V battery is connected to two resistors in series. One resistor is 5 ohms, and the other resistor is 10 ohms.
To Find :
a.) What is the total resistance for the circuit?
b.) What is the current passing through each resistor?
c.) What is the power used (in kW) for the 5-ohm resistor?
d.) What is the power used (in kW) for the 10-ohm resistor?
e.) If the battery runs for 20 hours, and its costs $2.50/kW*hr, what will the total charge be?
Solution:
V = IR
P = VI = I²R = V²/R
two resistors in series
Hence total resistance for the circuit = 5 + 10 = 15 Ω
current passing through each resistor = 30/15 = 2A
power used (in kW) for the 5-ohm resistor = 2² * 5 / 1000 = 20/1000 kw
= 0.02 kW
power used (in kW) for the 10 -ohm resistor = 2² * 10 / 1000 =40/1000 kw
= 0.04 kW
Total power consumption = 0.06 kw
battery runs for 20 hours, and its costs $2.50/kW*hr, what will the total charge be
0.06 * 20 * 2.5 = $ 3
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