Consider a 10 k ohm resistor and an initially uncharged 100 mu F parallel plate

Question

Consider a 10 k ohm resistor and an initially uncharged 100 mu F parallel plate air capacitor connected in series with a 3.0 Volt battery and an open switch. At a time t = 0, the switch is closed. After a time (t) the capacitor has 60% of its maximum charge. What is that time t and the charge Q on the capacitor? t = ln(2/5) see and Q= 180 mu C t = ln (2/5) see and Q - 300 mu C. t = ln(5/2) sec and Q = 180 mu C t = ln(5/2) sec and Q = 300 mu C t = 1 sec and Q = 180 mu C t = 1 see and Q = 300 mu C In the circuit described in question (4) above, a material with dielectric constant K is inserted to fill the space between the capacitor's parallel plates. How do the maximum charge and the time constant change? no change in either the maximum charge or the time constant maximum charge decreases by factor 1/K, the time constant increases by factor K maximum charge decreases by factor 1/K. the time constant decreases by factor 1/K maximum charge increases by factor K, the time constant increases by factor K maximum charge increases by factor K, the time constant decreases by factor 1/K

Explanation / Answer

4)    Here, Q = Qo * ( 1 - e-t/RC)

=>   0.60 = ( 1 - e-t/1)

=>   t   =   ln(5/2)    sec

=>   Q   = 100 * 10-6 * 3 * 0.6

=> Q = 180 microC

=> option c) is correct .

5)      maximum charge increases by factor K , time constant increases by factor K .

=>    option d) is correct .

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