In an electrical circuit, two resistors of 2 Ω and 4 Ω are connected in series to a 6 V battery, find the heatdissipated by the 4 Ω resistor in 5 s

Sure, here are the steps to solve the problem:

Step 1: Find the total resistance in the circuit. Since the resistors are in series, you simply add their resistances together. So, R_total = R1 + R2 = 2 Ω + 4 Ω = 6 Ω.

Step 2: Use Ohm's law (V = IR) to find the total current in the circuit. Rearrange the formula to solve for I: I = V/R. Plug in the values: I = 6V / 6Ω = 1 A.

Step 3: Now, find the voltage across the 4 Ω resistor. Use Ohm's law again: V = IR. Plug in the values: V = 1A * 4Ω = 4V.

Step 4: Use Joule's law of electric heating, which states that the heat dissipated by a resistor is given by H = I^2Rt, where I is the current through the resistor, R is the resistance, and t is the time. Plug in the values: H = (1A)^2 * 4Ω * 5s = 20 Joules.

So, the heat dissipated by the 4 Ω resistor in 5 seconds is 20 Joules.