M. (a) (i) (use of V = IR) R total = (ohm) V = =.0 V (use of V = IR) R = 9.0/.0 = 9.0 Ω r = 9.0.0 6.0 =.0 Ω or use of (E = I(R + r)) 9.0 = (7 + r) r = 9.0 7.0 =.0 Ω (iii) (use of W = Vlt) W = 9.0.0 5 60 W = 700 J (iv) energy dissipated in internal resistance =.0 5 60 = 600 (J) percentage = 00 600/700 = % CE from part aii (b) internal resistance limits current hence can provide higher current or energy wasted in internal resistance/battery less energy wasted (with lower internal resistance) or charges quicker as current higher or less energy wasted or (lower internal resistance) means higher terminal pd/voltage as less pd across internal resistance or mention of lost volts [0] PhysicsAndMathsTutor.com
M. (a) (i) energy changed to electrical energy per unit charge/coulomb passing through [or electrical energy produced per coulomb or unit charge] [or pd when no current passes through/or open circuit] () I = =.5 A () (iii) (use of = I(R + r) gives) = V + Ir and 8 = 6 + Ir () substitution gives 8 6 =.5r () (and r = 0.8 Ω) 4 (b) (i) (use of P = I R gives) P R =.5.4 = 5 W [or P = VI gives P = 6.5 = 5 W] () (allow C.E. for value of I from (a)) P T = 5 + (.5 0.8) () = 0 (W) () (allow C.E. for values of P R and I) (iii) E = 5 60 = 600 J () (allow C.E. for value of P from (i) and P T from ) 4 [8] PhysicsAndMathsTutor.com
M. (i) (V = IR gives) = (0 + 0 + )I () I = = 0.9 A () (0.94 A) V PQ = (0.9 ) () =.6 V () (allow C.E. for incorrect I in (i)) [or V PQ = 0.9 60 =.6 V] (I = 0.94 A gives.6 V) [or V PQ = =.6 V (iii) (P A = I R gives) P A = (0.9) 0 =.08 () W () [or P A = ] (allow C.E. for incorrect I in (i) or incorrect V in ) (iv) (E = P A t gives) E =.08 0 () =.6 J () (allow C.E. for incorrect P A in (iii)) [8] M4. (a) (i) work (done)/energy (supplied) per unit charge (by battery) () (or pd across terminals when no current passing through cell or open circuit) when switch is closed a current flows (through the battery) () hence a pd/lost volts develops across the internal resistance () (b) (use of ε = V + Ir) I = 5.8/0 = 0.58 (A) () 6.0 = 5.8 + 0.58r () r = 0./0.58 = 0.4 (Ω) () PhysicsAndMathsTutor.com
(c) need large current/power to start the car () (or current too low) internal resistance limits the current/wastes power(or energy)/reduces terminal pd/increases lost volts () [8] M5. (a) (use of E = V + Ir) = V + 40 0.0095 () V = 8.0()V () (b) ρ = RA/I =.6 0 7.9 0 5 /0.75 () R =.7 0 7 () Ωm () [5] M6. (a) (i) work done (by the battery) per unit charge () or (electrical) energy per unit charge or pd/voltage when open circuit/no current the resistance of the materials within the battery () or hindrance to flow of charge in battery or loss of pd/voltage per unit current (b) (i) (use of E = V + Ir) = V + 800 0.005 () (working/equation needs to be shown) V = 4 = 8.0V () (use of P = I r) P = 800 0.005 () (working/equation needs to be shown) P = 00 () W () or J s 5 PhysicsAndMathsTutor.com 4
(c) car will probably not start () battery will not be able to provide enough current () or less current or lower terminal pd/voltage [9] M7. (a) mention of pd across internal resistance or energy loss in internal resistance or emf > V pd across internal resistance/lost volts increases with current or correct use of equation to demonstrate (b) (i) y intercept.5 V (± 0.0 V) identifies gradient as r or use of equation substitution to find gradient or substitution in equation r = 0.45 ± 0.0 Ω (c) (i) same intercept double gradient (must go through.5, 0.40 ±.5 squares) same intercept horizontal line (d) (i) (use of Q = lt) Q = 0.89 5 = C use of P = I r P = 0.89 0.45 P = 0.6 W [] M8. (a) battery has internal resistance () current passes through (this resistance) () work done/voltage lost, which reduces the value of the emf () PhysicsAndMathsTutor.com QWC 5
(b) (i) circuit diagram to show: two cells in series () two resistors, each labelled r () (use of P = IV gives).6 =.5 I () (I = 0.64 (A)) (use of = V + Ir gives).0 =.5 +0.64 r () () 0.5 =.8r and r = 0.9 Ω () [or R bulb =.5 /.6 =.9 (Ω) and.5 =.9 I gives I = 0.64 (A) lost volts = (.5) = 0.5 (V) i.e. 0.5 (V) per cell 0.5 = 0.64r and r = 0.9 Ω] 5 (c) = V + Ir gives V = Ir + (equation of straight line) () intercept on y-axis gives () gradient gives ( )r () [] M9. (a) V = Ir + () (b) straight line (within st quadrant) () negative gradient () (c) : intercept on voltage axis () r: gradient () [5] M0. (a) (i) 6.0 (Ω) () 4.5 (V) () (iii) (use of I = V/R) I = 4.5/6.0 = 0.75 (A) () current through cell A = 0.75/ = 0.75 (A) () (iv) charge = 0.75 00 = () C () PhysicsAndMathsTutor.com 6
(b) cells C and D will go flat first or A and B last longer () current/charge passing through cells C and D (per second) is double/more than that passing through A or B () energy given to charge passing through cells per second is double or more than in cells C and D () or in terms of power [9] M. (a) (i) electrical energy produced (in the battery) per unit charge () [or potential/voltage across terminals when there is no current] there is a current (through the battery) () voltage lost across the internal resistance () Max (b) (i) = V + Ir () labelled scales () correct plotting () best straight line () : intercept on y axis () = 9. (± 0.) V () r : ( ) gradient = = 4. Ω () (range 4.0 to 4.) 8 [0] PhysicsAndMathsTutor.com 7