Given the circuit, we have a battery of 64.0V and two resistors of resistance 34.0Ω and 60.0Ω connected in series.
In a series circuit, the total resistance (R_total) is given by the sum of the individual resistances. Therefore, R_total is equal to 34.0Ω + 60.0Ω, which results in 94.0Ω.
In order to find the current across each resistor (I₃₄.₀₀ and I₆₀.₀₀), we employ Ohm's law which states V = IR, where V is voltage, I is current, and R is resistance. Rearranging for current gives us I = V/R. Applying this by dividing the battery voltage (V = 64.0V) by the total resistance (R = 94.0Ω) gives us the current value.
Calculating with the given values, we find that I₃₄.₀₀ = I₆₀.₀₀ = 64.0V / 94.0Ω = 0.6808510638297872 A.
Next, we determine the power across each resistor (P₃₄.₀₀ and P₆₀.₀₀). The power (P) is given by P = IV, where V is the voltage. In a series circuit, the voltage across each resistor is given by V = IR. So the power can also be calculated as P = I²R.
Therefore, calculating power across the 34.0Ω resistor, P₃₄.₀₀ = (0.6808510638297872A)² * 34Ω = 23.148936170212764W.
Similarly, calculating the power across the 60.0Ω resistor, P₆₀.₀₀ = (0.6808510638297872A)² * 60Ω = 40.851063829787236W.
In conclusion, the values are: I₃₄.₀₀ = 0.6808510638297872 A, P₃₄.₀₀ = 23.148936170212764W, I₆₀.₀₀ = 0.6808510638297872 A, and P₆₀.₀₀ = 40.851063829787236W.