R3 R13 R2 R4 R3 R6 R1 R2 R7 R5 R4

Two resistors, R = 3.09 S and R2 = 5.27 2, are connected R1 in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current I through R and the potential difference V across R2. I = A. R2 V2 = V

Consider the circuit below. The battery has an open-circuit voltage of ɛ = 30 V and an internal resistance ofr = 1N. R2 = 18.00 N E = 30.00 V R1 9.00 N %3D ir = 1.00 N R3 10.00 N R5 8.00 N R4 = 10.00 N Hint a. Find the equivalent resistance of the circuit (not including the internal resistance). Reg b. Find the current out of the battery (do not ignore the internal resistance). Ibat A c. Find the potential drop across each resistor. o Across Rj and R2: AV = V. Across R3 and R4: AV = V. Across Rs: Д V — V. Across r: AV = V. (These drops should add up to the value of ɛ.) d. Find the currents through Rị and R2: IRI A IR2 A e. Find the currents through R3 and R4: IR3 А IRA A f. What is the percentage of the total power supplied by the battery that is dissipated in the internal resistance of the battery? A total of % of the total power supplied by the battery is dissipated in the internal resistance r. 十

Three resistors, R1= 11 Ω, R2 = 50 Ω, and R3=R, are connected in series with a 24.0 V battery. The total current flowing through the battery is 0.20A.   A)Find the value of resistance R. Express your answer using two significant figures.   B) Find the potential difference across each resistor. Enter your answers numerically separated by commas. Express your answers using two significant figures.   C) If the voltage of the battery had been greater than 24.0 V, would your answer to part A have been larger or smaller? Explain