Two resistors are connected in series with an 34.76 volt battery. The resistors have values of R1 = 1.84 ohms and R2 = 9.71 ohms. What is the current (in amps) in this circuit? Round your answer to two decimal places.

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 with values 1.8 Ω, 2.6 Ω, and 4.2 Ω are connected in parallel in a circuit with a 7.8 V battery. Part A What is the total equivalent resistance? Express your answer using two significant figures.   Part B What is the voltage across each resistor? Express your answer using two significant figures.   Part C What is the power delivered to the 4.2 Ω resistor? Express your answer using two significant figures.

The current in a series circuit is 21.2 A. When an additional 7.38-Ω resistor is inserted in series, the current drops to 12.2 A. What is the resistance in the original circuit?