Part B The move is opened at . T= 0 s at what time has actually thecharge ~ above the capacitor decreased to 29% the its initialvalue? The principles used to resolve this problem are mix of collection resistance, Ohm’s law and also expression of fee on the capacitor, and the fee on the capacitor after ~ time ttt .

You are watching: The switch in the figure has been open for a long time. it is closed at t = 0s.

First calculation the indistinguishable resistance and also after that calculate the present in the circuit, next calculate the potential difference, lastly calculate the charge throughout the capacitor, in part (B) calculate the time required to charge ~ above the capacitor decrease to 29%29\% 29% the its early stage value.

The equivalent resistance of series resistance is together follows.

Req=R1+R2R_\rmeq = R_1 + R_2Req​=R1​+R2​

Here R1R_1R1​ and R2R_2R2​ room two resistances associated in series.

From Ohm’s law,

V=IRV = IRV=IR

Here, III is current, RRR is resistance, and VVV is potential difference.

The charge across the capacitor is,

Q=CVQ = CVQ=CV

Here QQQ is charge, CCC is capacitance and also VVV is potential difference.

The charge on the capacitor after ~ time ttt is as follows,

Q=Qmaxe−tRCQ = Q_\rmmaxe^\frac - tRCQ=Qmax​eRC−t​

Here, QmaxQ_\rmmaxQmax​ is best charge, RRR is resistance, ttt is time taken by capacitor, CCC is capacitance.

As the switch closed for long time the capacitor behaves like open up circuit, so over there is no present flow from the capacitor.

The listed below image shows the condition of circuit in ~ t=∞t = \infty t=∞ . Calculate the identical resistance by using series resistance formula.

The identical resistance of series resistance is,

Req=R1+R2R_\rmeq = R_1 + R_2Req​=R1​+R2​

Here R1R_1R1​ and also R2R_2R2​ room two resistances linked in series.

Substitute 60Ω60\rm \Omega 60Ω for R1R_1R1​ and also 40Ω40\rm \Omega 40Ω for R2R_2R2​ .

Req=60Ω+40Ω=100Ω\beginarrayc\\R_\rmeq = 60\rm \Omega + 40\rm \Omega \\\\ = 100\rm \Omega \\\endarrayReq​=60Ω+40Ω=100Ω​

Now calculation the current by using Ohm’s law.

From Ohm’s law,

V=IRV = IRV=IR

Here, III is current, RRR is resistance, and also VVV is potential difference.

Rearrange the equation V=IRV = IRV=IR for existing III .

I=VRI = \fracVRI=RV​

Substitute 100.0V100.0\rm V100.0V for VVV , and 100Ω100\rm \Omega 100Ω for ReqR_\rmeqReq​ .

I=100.0V100Ω=1.0A\beginarrayc\\I = \frac100.0\rm V100\rm \Omega \\\\ = 1.0\rm A\\\endarrayI=100Ω100.0V​=1.0A​

Now calculate the voltage throughout the 40Ω40\rm \Omega 40Ω resistor.

V′=I(40Ω)=(1.0A)(40Ω)=40V\beginarrayc\\V' = I\left( 40\rm \Omega \right)\\\\ = \left( 1.0\rm A \right)\left( 40\rm \Omega \right)\\\\ = 40\rm V\\\endarrayV′=I(40Ω)=(1.0A)(40Ω)=40V​

Now the maximum on the capacitor is,

The charge across the capacitor is,

Qmax=CVQ_\rmmax = CVQmax​=CV

Here QmaxQ_\rmmaxQmax​ is best charge, CCC is capacitance and also VVV is potential difference.

Substitute 2.0μF2.0\rm \mu F2.0μF for CCC and 40V40\rm V40V for V′V'V′ .

Qmax=(2.0μF)(40V)=80.0μC\beginarrayc\\Q_\rmmax = \left( 2.0\rm \mu F \right)\left( 40\rm V \right)\\\\ = 80.0\rm \mu C\\\endarrayQmax​=(2.0μF)(40V)=80.0μC​

The below image reflects the condition of circuit at t=0t = 0t=0 . Calculate the tantamount resistance t=0t = 0t=0 as soon as switch open.

The indistinguishable resistance of series resistance is,

Req=R1+R2R_\rmeq = R_1 + R_2Req​=R1​+R2​

Here R1R_1R1​ and R2R_2R2​ room two resistances linked in series.

Substitute 10Ω10\rm \Omega 10Ω because that R1R_1R1​ and 40Ω40\rm \Omega 40Ω for R2R_2R2​ .

Req=10Ω+40Ω=50Ω\beginarrayc\\R_\rmeq = 10\rm \Omega + 40\rm \Omega \\\\ = 50\rm \Omega \\\endarrayReq​=10Ω+40Ω=50Ω​

Now calculation the charge on capacitor after ~ time ttt when the fee on the capacitor lessened to 29%29\rm\% 29% of the early value.

The charge on the capacitor after ~ time ttt is,

Q=Qmaxe−tRCQ = Q_\rmmaxe^\frac - tRCQ=Qmax​eRC−t​

Here, QmaxQ_\rmmaxQmax​ is maximum charge, RRR is resistance, ttt is time bring away by capacitor, CCC is capacitance.

Rearrange the equation Q=Qmaxe−tRCQ = Q_\rmmaxe^\frac - tRCQ=Qmax​eRC−t​ for ttt .

t=−(RC)ln⁡(QQmax)t = - \left( RC \right)\ln \left( \fracQQ_\rmmax \right)t=−(RC)ln(Qmax​Q​)

Substitute 50Ω50\rm \Omega 50Ω because that RRR , 2.0μF2.0\rm \mu F2.0μF because that CCC , 0.29Qmax0.29Q_\rmmax0.29Qmax​ because that QQQ .

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t=−((50Ω)(2.0μF))ln⁡(0.29QmaxQmax)=−((50Ω)(2.0μF)(10−6F1F))ln⁡(0.29)=−((50Ω)(2.0×10−6F))(−1.23)=1.24×10−4s\beginarrayc\\t = - \left( \left( 50\rm \Omega \right)\left( 2.0\rm \mu F \right) \right)\ln \left( \frac0.29Q_\rmmaxQ_\rmmax \right)\\\\ = - \left( \left( 50\rm \Omega \right)\left( 2.0\rm \mu F \right)\left( \frac10^ - 6\rm F1\rm F \right) \right)\ln \left( 0.29 \right)\\\\ = - \left( \left( 50\rm \Omega \right)\left( 2.0 \times 10^ - 6\rm F \right) \right)\left( - 1.23 \right)\\\\ = 1.24 \times 10^ - 4\rm s\\\endarrayt=−((50Ω)(2.0μF))ln(Qmax​0.29Qmax​​)=−((50Ω)(2.0μF)(1F10−6F​))ln(0.29)=−((50Ω)(2.0×10−6F))(−1.23)=1.24×10−4s​

Convert second in to millisecond.

t=(1.24×10−4s)(103ms1s)=0.124ms\beginarrayc\\t = \left( 1.24 \times 10^ - 4\rm s \right)\left( \frac10^3\rm ms1\rm s \right)\\\\ = 0.124\rm ms\\\endarrayt=(1.24×10−4s)(1s103ms​)=0.124ms​