1. The current in a 44 resistor is 0.10 A. This resistor is in series with a28 r
ID: 1722206 • Letter: 1
Question
1. The current in a 44 resistor is 0.10 A. This resistor is in series with a28 resistor, and the series combination isconnected across a battery. What is the battery voltage? 2. Two resistors, 42.0 and 72.0 , are connected in parallel. Thecurrent through the second resistor is 4.30 A.(a) What is the current in the first resistor?
(b) What is the total power supplied to the tworesistors?
1. The current in a 44 resistor is 0.10 A. This resistor is in series with a28 resistor, and the series combination isconnected across a battery. What is the battery voltage? 2. Two resistors, 42.0 and 72.0 , are connected in parallel. Thecurrent through the second resistor is 4.30 A.
(a) What is the current in the first resistor?
(b) What is the total power supplied to the tworesistors?
(a) What is the current in the first resistor?
(b) What is the total power supplied to the tworesistors?
Explanation / Answer
Battery Voltage is equal to V = (44+28)*0.1A =7.2V
The total current I = I72 + I42 and I 72 = (42/42+72)I Then I =4.3A*114/42 =11.67A Then the current through the first resistor is I42= 11.67A-4.3A = 7.37A
Total power supplied to the two resistors is P =(4.3A)2*72 +(7.37A)2*42 =3612.5W
Hope this helps you.
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