Part B The switch is opened at . T= 0 s in ~ what time has actually thecharge on the capacitor lessened to 29% that its initialvalue?

The principles used to fix this problem are mix of collection resistance, Ohm’s law and also expression of charge on the capacitor, and the fee on the capacitor after time ttt .

You are watching: The switch in the figure(figure 1) has been closed for a very long time.

First calculation the indistinguishable resistance and after that calculate the existing in the circuit, following calculate the potential difference, lastly calculate the charge throughout the capacitor, in part (B) calculate the time required to charge on the capacitor decrease come 29%29\% 29% the its early value.

The tantamount resistance of series resistance is as follows.

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

Here R1R_1R1 and also R2R_2R2 are 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 fee on the capacitor ~ time ttt is together follows,

Q=Qmaxe−tRCQ = Q_\rmmaxe^\frac - tRCQ=QmaxeRC−t

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

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

The below image mirrors the problem of circuit in ~ t=∞t = \infty t=∞ .

Calculate the identical resistance by using series resistance formula.

The indistinguishable resistance of series resistance is,

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

Here R1R_1R1 and also R2R_2R2 space two resistances connected in series.

Substitute 60Ω60\rm \Omega 60Ω because that R1R_1R1 and 40Ω40\rm \Omega 40Ω because that 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 existing by utilizing Ohm’s law.

From Ohm’s law,

V=IRV = IRV=IR

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

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

I=VRI = \fracVRI=RV

Substitute 100.0V100.0\rm V100.0V because that VVV , and also 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 preferably on the capacitor is,

The charge throughout the capacitor is,

Qmax=CVQ_\rmmax = CVQmax=CV

Here QmaxQ_\rmmaxQmax is maximum charge, CCC is capacitance and VVV is potential difference.

Substitute 2.0μF2.0\rm \mu F2.0μF for CCC and also 40V40\rm V40V because that 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 mirrors the condition of circuit at t=0t = 0t=0 .

Calculate the indistinguishable resistance t=0t = 0t=0 when 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 are two resistances connected in series.

Substitute 10Ω10\rm \Omega 10Ω because that R1R_1R1 and also 40Ω40\rm \Omega 40Ω because that 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 charge on the capacitor lessened to 29%29\rm\% 29% of the early value.

The fee on the capacitor ~ time ttt is,

Q=Qmaxe−tRCQ = Q_\rmmaxe^\frac - tRCQ=QmaxeRC−t

Here, QmaxQ_\rmmaxQmax is preferably charge, RRR is resistance, ttt is time take away by capacitor, CCC is capacitance.

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

t=−(RC)ln(QQmax)t = - \left( RC \right)\ln \left( \fracQQ_\rmmax \right)t=−(RC)ln(QmaxQ)

Substitute 50Ω50\rm \Omega 50Ω because that RRR , 2.0μF2.0\rm \mu F2.0μF for 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(Qmax0.29Qmax)=−((50Ω)(2.0μF)(1F10−6F))ln(0.29)=−((50Ω)(2.0×10−6F))(−1.23)=1.24×10−4s

Convert second in come 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