5. An energy source forces a constant current of 2 A for 10 s to flow through a lightbulb. If 2.3 kJ is given off in the form of light and heat energy, calculate the voltage drop across the bulb. 6. To move charge q from point a to point b requires -30J . Find the voltage drop v_(ab) if: (a) q=2C, (b) q=-6C 7. Find the power delivered to an element at t=5 ms if the current is i=5cos60pi t A and the voltage is : (a) v=2iV, (b) V:(10+5int _(0)^tidt)v

5. To calculate the voltage drop across the bulb, we can use the formula:Work done (W) = Voltage (V) * Current (I) * Time (t)Given that the energy given off in the form of light and heat is 2.3 kJ (or 2300 J), the current is 2 A, and the time is 10 s, we can rearrange the formula to solve for voltage:Voltage (V) = Work done (W) / (Current (I) * Time (t))Plugging in the values:Voltage (V) = 2300 J / (2 A * 10 s)Voltage (V) = 115 VTherefore, the voltage drop across the bulb is 115 V.6. To find the voltage drop , we can use the formula:Voltage (V) = Work done (W) / Charge (q)Given that the work done is -30 J, we can calculate the voltage drop for both cases:(a) For :Voltage (V) = -30 J / 2 CVoltage (V) = -15 VTherefore, the voltage drop is -15 V.(b) For :Voltage (V) = -30 J / -6 CVoltage (V) = 5 VTherefore, the voltage drop is 5 V.7. To find the power delivered to the element at ms, we need to calculate the current and voltage at that specific time.(a) For :Current (I) = 5 cos(60π * 5 ms) ACurrent (I) = 5 cos(300π) A (I) = 5 AVoltage (V) = 2 * 5 VVoltage (V) = 10 VPower (P) = Voltage (V) * Current (I)Power (P) = 10 V * 5 APower (P) = 50 WTherefore, the power delivered to the element at ms is 50 W.(b) For :Current (I) = 5 cos(60π * 5 ms) ACurrent (I) = 5 cos(300π) ACurrent (I) = 5 AVoltage (V) = (10 + 5 * ∫₀⁵ms I dt) VVoltage (V) = (10 + 5 * ∫₀⁵ms 5 dt) VVoltage (V) = (10 + 5 * 5 * 5 ms) VVoltage (V) = (10 + 125) VVoltage (V) = 135 VPower (P) = Voltage (V) * Current (I)Power (P) = 135 V * 5 APower (P) = 675 WTherefore, the power delivered to the element at ms is 675 W.