Electrical Circuits (DC)

loop.gif (706 bytes)
loopstretch.gif (1963 bytes)
Kirchhoff's Loop Rule:  The algebraic sum of potential drops along a closed loop is equal to zero.

In the image to the left the potential (V) is analogous to height.  According to the Loop Rule the potential difference caused by the battery (varepsilon.gif (50 bytes)) must be compensated for by the potential drops across the two resistors (r and R).  Notice that the potential (V) starts at Va and then returns again to Va after resistor R.

Previously Asked Questions

Q:    "A battery has an initial internal resistance of 0.75 omega.gif (841 bytes) and an emf of 9 V.  It is placed a cross a 5 omega.gif (841 bytes) resistor and a 10 µF capacitor hooked up in parallel.
a)    After a the capacitor has charged, what is the current through the resistor?
b)    What is the charge on the capacitor?
c)    If the battery is disconnected, how long will it take the capacitor to reach one-third of its initial voltage?"

A:    First we will start off by drawing a picture of the circuit.

circuit1.gif (2464 bytes)a)  After a the capacitor has charged, what is the current through the resistor?
After the capacitor has charged, there is no current passing through it.  This means the resistors (the internal resistance of the battery, r, and the resistor, R ) are connected in series.  Thus, the equivalent resistance of the circuit, Req , is the sum of the internal resistance of the battery, r, and the resistance of the resistor, R:

Req = r + R


Because the resistors are in series, the current in this circuit, I, is constant and equal to the emf, varepsilon.gif (50 bytes), over the equivalent resistance, Req:

Ivarepsilon.gif (50 bytes) / Req

Now solve for I:
I   = varepsilon.gif (50 bytes) / Req
    = 9 V / (.75
omega.gif (841 bytes) + 5 omega.gif (841 bytes))
    = 1.57 A

The current through the battery is 1.57 A.

b)  What is the charge on the capacitor?
After the capacitor, C, is charged, the voltage across it is the same as the voltage across the resistor, R, (= IR).  Therefore, the charge, Q, on the capacitor, C, is:

Q = CVR = C (IR) = 10 µF (1.57 A) (5 omega.gif (841 bytes)) = 78.3 µC.

The charge on the capacitor is 78.3 µC.

c)  If the battery is disconnected, how long will it take the capacitor to reach one-third of its initial voltage?circuit2.gif (1581 bytes)
In our picture, we switch the battery off with switch, S.  With the switch opened, the battery and its internal resistance drop out of the circuit.  We can then draw a new circuit, with the capacitor as the source of power.

Since the capacitor is discharging, the charge on it varies with time according to the equation q(t) = Q e (- t / RC).  To find the time, t, it takes the capacitor to reach one-third of its initial capacity, we substitute
q(t) = Q/3 into the equation and solve for t:

q(t) = Q e (- t / RC)

Q/3 = Q e (- t / RC)

1/3 = e (- t / RC)

ln (1/3) = ln (e (- t / RC))

- ln 3 = - t / RC

t = ( ln 3 ) (RC)

t = (ln 3) (5 omega.gif (841 bytes)) (10 µF)

t = 54.9 µsec

After 54.9 µsec, the capacitor has be one-third discharged.

Q:    If it is the current flowing through your body that determines the amount of damage done from a shock, why do the signs warn of high voltage and not high current?

A:    According to Ohm's Law ( I = V/R ), the current is proportional to voltage for a constant resistance.  Therefore, with the resistance of the human body given, the larger the voltage the larger the current passing through the body.   Contact with a high voltage source is thus more dangerous than with a low voltage source.

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The Loop Rule:  The algebraic sum of the charges in potential encountered in a complete traversal of any loop of a circuit must be zero.

The Junction Rule: The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction.


Emf (varepsilon.gif (50 bytes)) 28-1.gif (169 bytes)
Single-loop circuits 28-4.gif (141 bytes)
Power P = iV
Rate of energy transfer to thermal energy within a battery Pr = i2 r
Rate of chemical energy change within a battery Pemf = i varepsilon.gif (50 bytes)
Series resistances 28-7.gif (206 bytes)
Parallel Resistances 28-21.gif (241 bytes)
Charging capacitor q = C varepsilon.gif (50 bytes) ( 1 - e - t/RC )
Current during charging of the capacitor 28-31.gif (310 bytes)
Discharging capacitor q = q0 e - t/RC
Current during discharging of the capacitor 28-37.gif (344 bytes)

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List of Topics

Measurements Electric Potential Magnetism Electrical Circuits (AC) Optical Instruments: Mirrors and Lenses
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Electric Fields Current and Resistance Magnetism in Matter Electromagnetic Waves Diffraction
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