**Procedure:**1. Suppose the DC power source is set to 15 volts.2. In Figure 4, it shows a 10 μF capacitor connected with a 470 kΩ resistor, and the switch is set to position B.3. Calculate the time it will take for the capacitor to reach 63.2% of full charge when the switch is changed to position A. One time constant (TC) equals: __________ seconds.**Figure 4 Explanation:**- The circuit diagram includes a variable DC power supply (0-20 V), an ammeter (A), a 10 μF capacitor, and resistors R1 (470 kΩ), R2 (10 kΩ).- The switch can toggle between positions A and B, affecting the circuit configuration.4. Calculate the total time needed for the capacitor to reach full charge when the switch is changed from position B to A: __________.5. Calculate the voltage that should be across the capacitor for each time constant, from one through five, when the switch is connected to position A. **Record** on the lines below. - 1 TC: __________ - 2 TC: __________ - 3 TC: __________ - 4 TC: __________ - 5 TC: __________ 6. Change the switch position from A to B. Calculate the time it will take for the capacitor to reach 63.2% of full discharge. One TC equals: __________ seconds.7. Calculate the total time needed for the capacitor to reach the full discharge voltage when the switch is changed from position A to B: __________.8. Calculate the voltage that should be across the capacitor for each time constant, one through five, when the switch is changed from position A to B. **Record** on the lines below. Capacitor voltage after: - 1 TC: __________ - 2 TC: __________ - 3 TC: __________ - 4 TC: __________ - 5 TC: __________

Fig: Given circuit Procedure 1: Procedure 2: Procedure 3: In the above circuit capacitor is charging through resistance by the voltage source of 15V Given that V(t)= 0.63 Vs Although for RC circuit fomula for voltage across charging capacitor is given by V(t)=Vs1-e-tτwhere,τ=time constant=RC Hence,0.63Vs=Vs1-e-tτ0.63=1-e-tτ-0.37=-e-tτ0.37=e-tτln(0.37)=ln(e-tτ)-0.994=-tτt=τ (one time constant)τ=RCτ=470×103×10×10-6 secτ=4.7 sec. Hence 4.7 sec it will take for the given capacitor to reach 63.3 % of the full charge. Procedure 4: Although When capacitor is charging from voltage source time take by capacitor to reach 100% of the source voltage is equivalent to the 5 times of the time constant. Hence t100%=5τt100%=5×4.7 sect100%=23.50 sec Procedure 5: Although voltage across charging capacitor for different time constant is given by Time constant voltage across capacitor τ 63.2% of the source voltage 2τ 86.5% of the source voltage 3τ 95.0% of the source voltage 4τ 98.2% of the source voltage 5τ 99.30% of the source voltage To calculate the voltage across capacitor for different time constant Hence for given circuit voltage across charging capacitor at different time constant is Time constant voltage across capacitor τ 0.63×15=9.45V 2τ 0.86×15=12.90V 3τ 0.95×15=14.25V 4τ 0.98×15=14.7V 5τ 0.99×15V≅15VProcedure 6: Although for the above circuit capacitor is discharging through resistance and for discharging capacitor time taken by the capacitor to reach 63.20% of the full discharge is equivalent to the one time constant Hence t=τt=4.7 sec. Procedure: 7 Although total discharging time of the capacitor is t=5τt=5×4.7 sec.t=23.5 sec Procedure 8: Although voltage across discharging capacitor for different time constant is given by Time constant voltage across capacitor τ 36.6% of the source voltage 2τ 13.5% of the source voltage 3τ 5% of the source voltage 4τ 1.8% of the source voltage 5τ 0.7% of the source voltage To calculate the voltage across capacitor for different time constant Hence for given circuit voltage across discharging capacitor at different time constant is Time constant voltage across capacitor τ 0.36×15=5.4V 2τ 0.13×15=1.95V 3τ 0.05×15=0.75V 4τ 0.018×15=0.27V 5τ 0.007×15V≅0V