Final answer:
The voltages across the two resistors placed in series with resistances of 40Ω and 60Ω, when 5V is supplied, are 2V for the 40Ω resistor and 3V for the 60Ω resistor.
Step-by-step explanation:
When 5V is supplied to a circuit of two resistors placed in series, with resistances of 40Ω and 60Ω, we need to calculate the voltage across each resistor. To find the voltages, we use Ohm's law, which states that V = IR, where 'V' is voltage, 'I' is current, and 'R' is resistance. Because the resistors are in series, the current through them is the same. The total resistance of the series circuit is the sum of the individual resistances, which is 40Ω + 60Ω = 100Ω. The current in the circuit is therefore I = V / R = 5V / 100Ω = 0.05A.
The voltage drop across each resistor can then be calculated by multiplying the current by the resistance of each resistor. For the 40Ω resistor, the voltage across is V1 = I * R1 = 0.05A * 40Ω = 2V. Similarly, for the 60Ω resistor, the voltage across is V2 = I * R2 = 0.05A * 60Ω = 3V. Therefore, the voltage across the 40Ω and 60Ω resistors are 2V and 3V respectively.