5. How much work is done in moving a charge of 5 C across to a point having a p.d. [Ans: 100 D ] in a house, an electric bulb of 60 W is used for 15 hours and an electric heater of 750 W is used for 10 hours evenc dav. Determine the cost of using the bulb and

Let's break down the problem step by step. ### Part 1: Work Done in Moving a Charge 1. **Known Conditions:** - Charge (\(Q\)) = 5 C - Potential difference (\(V\)) = 20 V (since the answer is given as 100 D, we can deduce that \(V = 20\) V because Work done \(W = Q \times V\)) 2. **Formula for Work Done:** \[ W = Q \times V \] 3. **Calculating Work Done:** \[ W = 5 \, \text{C} \times 20 \, \text{V} = 100 \, \text{J} \] Thus, the work done in moving the charge is \(100 \, \text{J}\). ### Part 2: Cost of Using the Bulb and Heater 1. **Known Conditions:** - Power of the bulb (\(P_b\)) = 60 W - Time used for the bulb (\(t_b\)) = 15 hours - Power of the heater (\(P_h\)) = 750 W - Time used for the heater (\(t_h\)) = 10 hours - Cost per kWh = \(C\) (we will assume a value for calculation, e.g., \(C = 0.15\) per kWh) 2. **Calculating Energy Consumption:** - Energy consumed by the bulb: \[ E_b = P_b \times t_b = 60 \, \text{W} \times 15 \, \text{h} = 900 \, \text{Wh} = 0.9 \, \text{kWh} \] - Energy consumed by the heater: \[ E_h = P_h \times t_h = 750 \, \text{W} \times 10 \, \text{h} = 7500 \, \text{Wh} = 7.5 \, \text{kWh} \] 3. **Total Energy Consumption:** \[ E_{total} = E_b + E_h = 0.9 \, \text{kWh} + 7.5 \, \text{kWh} = 8.4 \, \text{kWh} \] 4. **Calculating Cost:** \[ \text{Cost} = E_{total} \times C = 8.4 \, \text{kWh} \times 0.15 \, \text{per kWh} = 1.26 \] Thus, the total cost of using the bulb and heater is approximately \(1.26\). ### Summary of Results - Work done in moving the charge: \(100 \, \text{J}\) - Cost of using the bulb and heater: \(1.26\) (assuming cost per kWh is \(0.15\))