show working out please

4} Having reached sv stored the same capacitor is now discharged through a 22kt]! resistor. The Discha rging equation for this circuit is: i; 1:? = Ve T where: 1 = CH and is called the time constant 1:: =the voltage across the capacitor i": =the time since the supply voltage applied in seconds E =the capacitance of the capacitor in Farads R =the resistance of the resistor in ohms a} Calculate the time constant for the circuit. h] F'lot the discharging curve of vertical]: against horizontal] over the range III to El] milliseconds. c} From your graph use a tangent to the curve to estimate the gradient when i: 2 15m. cl] Differentiate the discharging equation with respect to [and use this to calculate the rate of change of voltage when i": = T. Ems. e} Compare your estimated and calculated answers. is there any difference between the values. why. what could he done to reduce errors? {Hint ensure your gradient and rate of change have the same units]- For Part 4) 1. Time constant calculated 2. Table of values calculated to create graph of i(vertical) against t (horizontal) 3. Graph of v(vertical) against t(horizontal) plotted on graph paper. 4. Tangent to graph at t=7.5ms drawn appropriately 5. Written working of gradient of tangent determined using appropriate methods 6. Written working of discharging equation differentiated correctly 7. Written working of value for rate of change at t=7.5ms determined correctly using differentiated discharging equation. 8. Comparison of calculated and estimated rates of change