Two batteries are in a circuit with 3 resitors. The circuit is in the shape of a
ID: 2056390 • Letter: T
Question
Two batteries are in a circuit with 3 resitors. The circuit is in the shape of a rectangle, with a 9V battery on the left side and a 7V battery on the right side. At the top there are two resistors in a series, the one on the left is 10 and the one on the right is 14. At the bottom of the square, there is a third resistor of 20.
a) What is the current out of the 9V battery?
b) What is the power dissipated by the 14 resistor?
Could you please explain the steps involved and all variables? Thanks so much : )
Explanation / Answer
Your whole question is understood only for the fact that are the battery connected with positive terminal to the positive terminal or they are connected in reverse direction current value very much depends on that I am solving assuming positive is connected to positive of the battery, if it is opposite the equivalent battery just do = V1 + V2 now I will use V1 - V2 = 9-7 = 2 V total resistance = 10+14 + 20 = 44 Ohms { all in series} so current out of 9 V battery = 2/44 = 1/11 A b) Power dissipated in 14 Ohms resistor = i^2 * 14 = 1/121 * 14 = .1157 W