You must remember the following equations to make you strong in answering multiple choice questions involving *electric circuits*:

**(1)** Electric current, ** I = nAvq** where

*n*is the number density (number per unit volume) of mobile charge carriers,

*A*is the area of cross section of the conductor,

*v*is the drift velocity of the charge carriers and

*q*is the charge on each carrier.

Since the charge carriers in a conductor are electrons of charge *e*, the above expression becomes

*I = nAve*

**(2)** Resistivity **ρ = RA/L**

*where*

*R*,

*A*and

*L*are respectively the resistance, area of cross section and the length of the conductor.

**(3) **If a wire of resistance *R* is stretched so that its length becomes *n* times its original length, its area of cross section becomes (1/*n*) times its original area of cross section. The resistance of the stretched wire therefore becomes* n*

**.**

^{2}*R*[The stretch can be described also in terms of the radius of the wire. Thus if a wire of resistance *R* is stretched so that its radius becomes (1/*n*) times the original radius, its area of cross section becomes (1/*n*^{2}) times the original area of cross section and its length becomes *n*^{2 }times the original length. The resistance of the wire therefore becomes ** n^{4}R**].

**(4) **Conductivity *s** * is the reciprocal of resistivity: *s** *** = 1/ρ**

**(5)** Effective value (R) of resistances in series: *R =R _{1}*

**+**

*R*_{2 }+*R*_{3}+*R*_{4}+….etc.**(6) **Effective value (*R*) of resistances in parallel is given by the reciprocal relation,

**1/ R = 1/R_{1} + 1/R_{2} + 1/R_{3} +… etc.**

**(7**) **(a)** The equivalent emf of a series combination of *n* cells is the sum of their individual emf’s:

**Ɛ*** ***=** **Ɛ**_{1}** +** **Ɛ**_{2}** +**** ****Ɛ**_{3}** +**** ****Ɛ**_{4}** +……etc**

** (b****)** The equivalent internal resistance of a series combination of n cells is the sum of their internal resistances:

*r ***= ***r*_{1}_{ }*+ r*_{2 }*+ r*_{3 }*+ r*_{4 }*+***……etc**

**(8) **If there an n cells of emf Ɛ_{1}, Ɛ_{2}, Ɛ_{3}, …… Ɛ_{n}, and of internal resistances *r*_{1}*,*_{ }* r*_{2 }*, ** r*_{3}*, ** **r** _{4}*…….

*r*

_{n}

_{ }*respectively, connected in parallel, the combination is equivalent to a single cell of emf Ɛ*

_{ }_{eq}

*and internal resistance*

*r*

*, such that*

_{eq} **1/***r*_{eq}** = 1/***r*_{1}** +1/***r*_{2}** +1/***r*_{3}** +……..+1/***r*_{n}** **and

** ****Ɛ**_{eq }**/***r*_{eq}** = ****Ɛ**_{1}**/***r*_{1}** +**** ****Ɛ**_{2}**/***r*_{2}** +**** ****Ɛ**_{3}**/***r*_{3}** +…….+****Ɛ**_{n}** ****/***r*_{n}

**(9) Kirchhoff’s rules**:

**(i)** *Junction rule**: **At any junction, the sum of the currents entering the junction is equal to the** **sum of currents leaving the junction**.*

**(ii)*** **Loop** rule**: The algebraic sum of changes in** **potential around any closed loop** **involving** **resistors and cells is zero**.*

**(10)** **Power **in a D.C. circuit = ** VI** =

*I*

^{2}

*R***=**

*V*

^{2}**/**where

*R**V*is the voltage,

*I*is the current and

*R*is the resistance.

**(11) **If** **devices consuming powers *P*_{1}, *P*_{2},* P*_{3},* P*_{4} ….etc. (at the same supply voltage *V*) are connected ** in parallel** across the supply voltage

*V*, the total power consumed by them is given by

*P ***= P_{1}+P_{2}+ P_{3}+ P_{4} +….etc.**

**(12) **If** **devices consuming powers *P*_{1}, *P*_{2},* P*_{3},* P*_{4} ….etc. (at the same supply voltage *V*) are connected ** in series** and the series combination is connected across the supply voltage

*V*, the total power consumed by them is given by the reciprocal relation,

**1/ P = 1/P_{1 }+1/P_{2 }+1/P_{3 }+1/P_{4} +….etc.**

**(13) **A cell will transfer maximum power to a load if the internal resistance of the cell is equal to the resistance of the load.

** Transients in RC circuits **are additionally included for the AP Physics C Examination. AP Physics C aspirants should therefore remember the following points also:

**(i) **When a capacitor of capacitance C farad is charged by connecting it in series with a resistance

*R*ohm and a battery of emf

*V*volts, the charge

*Q*coulomb on the capacitor after a time

*t*seconds

*is given by*

*Q =Q _{0 }*

**(1–**

**e**

^{–t/RC})

where *Q _{0}* = CV which is the final maximum charge (at infinite time) on the capacitor and ‘e’ is the base of natural logarithm. [

*Q = Q*exp(–t/RC)].

_{0}The product *RC* is the time constant of the RC circuit and has dimensions of time. If *R* is in ohm and *C* is in farad, *RC* is in seconds.

**(ii) **When a capacitor of capacitance C farad having initial charge *Q _{0}* coulomb is discharged through a resistance

*R*ohm, the charge

*Q*coulomb on the capacitor after a time

*t*seconds

*is given by*

*Q =Q _{0 }*

**e**

^{–t/RC}[Or, *Q = Q _{0}* exp(–t/RC)].

The** time constant** of the RC circuit can be defined using the charging process as well as the discharge process. It is the time required for the charge to build up to [1– (1/e)] times (which is 63.2%) the final maximum charge. It can be defined also as the time required for the charge to decay to (1/e) times (which is 36.8%) the initial charge.

[The growth and decay of *charge on a capacitor* *in a CR circuit* is similar to the growth and decay of *current through an inductor in an LR circuit*. If an inductor of inductance L and a resistor of resistance R are connected in series with a battery, the nature of increase of current *I* through the circuit is exponential with time and is given by

*I = I _{0 }*

**(1–**

**e**where

^{–Rt/L})*I*is the final (maximum) current in the circuit.

_{0}On disconnecting the battery, the current in the circuit decays exponentially with time as given by the equation

*I = I _{0 }*

**e**

^{–Rt/L}Here L/R is the ** time constant** of the LR circuit].

very helpful thanks--now i understand the equations better

ReplyDeleteHello alex,

ReplyDeleteVery happy to hear that the site is helpful to you. Best wishes.

very useful, site provides the concepts

ReplyDeleteThank you anonymous

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