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8:32 1 ..1 LTE D D 24:00 Exit Enter answer D5. Using the graph below, as well as the fact that the resistance in the circuit

i need help with this one that's all i have
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Answer #1

You have not provided the graph labels but for capacitance discharging we expect this sort of graph between time t and log of voltage i.e ln V because V = Vexpl-t/RC)

Hence In(V) = In (V.) – t/RO

where RC is time constant, hence graph between ln V and t is straight line with slope -1/RC

FROM YOUR GRAPH slope looks like -0.0875 (points are not clear hence might be bit different, I used points (8,-0.85) and (12,-1.2))

Hence -0.0875=-1/RC

given R =100 ohm

Hence C=1/8.75 F

C =0.114 F

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