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:<br /><br />Work done (W) = Voltage (V) * Current (I) * Time (t)<br /><br />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:<br /><br />Voltage (V) = Work done (W) / (Current (I) * Time (t))<br /><br />Plugging in the values:<br /><br />Voltage (V) = 2300 J / (2 A * 10 s)<br /><br />Voltage (V) = 115 V<br /><br />Therefore, the voltage drop across the bulb is 115 V.<br /><br />6. To find the voltage drop $v_{ab}$, we can use the formula:<br /><br />Voltage (V) = Work done (W) / Charge (q)<br /><br />Given that the work done is -30 J, we can calculate the voltage drop for both cases:<br /><br />(a) For $q=2C$:<br /><br />Voltage (V) = -30 J / 2 C<br /><br />Voltage (V) = -15 V<br /><br />Therefore, the voltage drop $v_{ab}$ is -15 V.<br /><br />(b) For $q=-6C$:<br /><br />Voltage (V) = -30 J / -6 C<br /><br />Voltage (V) = 5 V<br /><br />Therefore, the voltage drop $v_{ab}$ is 5 V.<br /><br />7. To find the power delivered to the element at $t=5$ ms, we need to calculate the current and voltage at that specific time.<br /><br />(a) For $v=2iV$:<br /><br />Current (I) = 5 cos(60π * 5 ms) A<br /><br />Current (I) = 5 cos(300π) A (I) = 5 A<br /><br />Voltage (V) = 2 * 5 V<br /><br />Voltage (V) = 10 V<br /><br />Power (P) = Voltage (V) * Current (I)<br /><br />Power (P) = 10 V * 5 A<br /><br />Power (P) = 50 W<br /><br />Therefore, the power delivered to the element at $t=5$ ms is 50 W.<br /><br />(b) For $V:(10+5\int _{0}^{t}idt)v$:<br /><br />Current (I) = 5 cos(60π * 5 ms) A<br /><br />Current (I) = 5 cos(300π) A<br /><br />Current (I) = 5 A<br /><br />Voltage (V) = (10 + 5 * ∫₀⁵ms I dt) V<br /><br />Voltage (V) = (10 + 5 * ∫₀⁵ms 5 dt) V<br /><br />Voltage (V) = (10 + 5 * 5 * 5 ms) V<br /><br />Voltage (V) = (10 + 125) V<br /><br />Voltage (V) = 135 V<br /><br />Power (P) = Voltage (V) * Current (I)<br /><br />Power (P) = 135 V * 5 A<br /><br />Power (P) = 675 W<br /><br />Therefore, the power delivered to the element at $t=5$ ms is 675 W.