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CBSE Class 10 Answered

How to calculate steady state current in a circuit with both resistors and capacitors when they are in    1. Series    2. Parallel .          Provided that the potential difference across battery is also given.  
Asked by acv27joy | 05 May, 2018, 10:41: PM
answered-by-expert Expert Answer
 
(1) Resistance and Capacitor are in series
 
Let us consider resistance R Ω and a capacitor of capacitance C Farads are connected in series and the capacitor is getting charged. 
 
when a small amount of charge dq is charged on the capacitor, we can write the energy conservation equation as
 
E×dq = i2×R×dt + d[q2/(2C)] ..............(1)
 
eqn.(1) can be written as E×dq = i2×R×dt + (q/C)dq ............. (2)
 
Let us divide eqn.(2) by dt, and cancel out the common term dq/dt also i ( i= dq/dt). Then eqn.(2) is rewritten as
 
E = i×R + (q/C) ..............(3)
 
now we substitute i = dq/dt, then 
 
E = (dq/dt)×R +(q/C); after some rearrangment we can write dq/(q-CE)  = -dt/(RC) ....(4)
 
by integrating (4) and using the ininitial condition q = 0 at t = 0, we have q(t) = CE{1 - exp[-t/(RC)]} .......(5)
 
if we differentiate and substitue dq(t)/dt = i(t), then we have i(t) = (E/R)exp(-t/RC) ............(6)
 
Eqn.(6) shows the current as a function of time. The product RC is termed as time constant.
Steady state will be reached after many time constants.
At starting point when t=0, current value E/R and after many time constants current drop down to zero.
------------------------------

 
When Resistor and capacitor are in parallel and connected to battery, the circuit is shown in above figure.
In this case capacitor is directly connected to battery and is quickly charged to the value CE, where E is EMF of battery.
Steady current E/R flows through the resistance
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