September 2011 - Career Point

(b) Real cellr,EiRise upr,EDrop upE – ir–E – ir(c) Electric resistanceRi R iDrop up–IR+IRiRise upWhen a battery being charged, the terminalvoltage is greater than its emf V = E + Ir.Kirchhoff's Law : Kirchhoff's law is able tosolve complicated circuit problems.(i) First Law : Incoming current = Outgoing currentI 1 + I 2 = I 3 + I 4 + I 5I 2I5I 1I 3I 4This law is based upon conservation principle ofcharge.(ii) Second Law : (Loop rule or voltage law.) Thislaw is based upon conservation principle ofenergy.Grouping of resistors :Case I : Resistors in seriesR MN = R eq = R 1 + R 2M R 1 R 2 NIn general,R eq = R 1 + R 2 + ... + R nCase II : Resistors in parallel1R MN=In general,1R MN=1R eqMR 1=1 1 +R 2R 1R 1R 2N1 1 1+ + ... +R 2 R nProblem solving strategy. : Power and Energy in circuitsStep 1 Identify the relevant concepts :The ideas of electric power input and output can beapplied to any electric circuit. In most cases you’llknow when these concepts are needed, because theproblem will ask you explicitly to consider power orenergy.Step 2 Set up the problem using the following steps :Make a drawing of the circuit.Identify the circuit elements, including sources ofemf and resistors.Determine the target variables. Typically theywill be the power input or output for each circuitelement, or the total amount of energy put into ortaken out of a circuit element in a given time.Step 3 Execute the solution as follows :A source of emf ε delivers power εI into a circuitwhen the current I runs through the source from –to +. The energy is converted from chemicalenergy in a battery, from mechanical energy in agenerator, or whatever. In this case the source hasa positive power output to the circuit or,equivalently, a negative power input to thesource.A source of emf power εI from a circuit – that is,it has a negative power output, or, equivalently, apositive power input–when currents passesthrough the source in the direction from + to –.This occurs in charging a storage battery, whenelectrical energy is converted back to chemicalenergy. In this case the source has a negativepower output to the circuit or, equivalently, apositive power input to the source.No matter what the direction of the currentthrough a resistor, It removes energy from acircuit at a rate given by VI = I 2 R = V 2 /R, whereV is the potential difference across the resistor.There is also a positive power input to the internalresistance r of a source, irrespective of thedirection of the current. The internal resistancealways removes energy from the circuit,converting it into heat at a rate I 2 r.You may need to calculated the total energydelivered to or extracted from a circuit element ina given amount of time. If integral is just theproduct of power and elapsed time.Step 4 Evaluate your answer : Check your results,including a check that energy is conserved. Thisconservation can be expressed in either of two forms:“net power input = net power output” or “thealgebraic sum of the power inputs to the circuitelements is zero.”Problem solving strategy : Series and ParallelStep 1 Identify the relevant concepts : Many resistornetworks are made up of resistors in series, inparallel, or a combination of the two. The keyconcept is such a network can be replaced by a singleequivalent resistor.XtraEdge for IIT-JEE 22 SEPTEMBER 2011