) CURRENT ELECTRICITY HOMEWORK SOLUTIONS. πd & A = 1. Given : 4 Solution : The resistance of a wire R is given by

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1 . CUNT CTICITY HOMWOK SOUTIONS. Given : Soution : m ρ. 0-7 Ωm Ω r? ρ ρ πr πr r r m r 0.6 mm. Given : ρ P ρ Q d P d Q P Q P Q? Formua : ρ πd & Soution : The resistance of a wire is given by P ρ P. P P P ρ P. P πd P P ρ P P πd P... (i) Simiary resistance of wire Q is, ρ Q Q Q... (ii) πd Q Dividing (i) by (ii) P ρ P P πd Q Q πd P ρ Q d Q P ρ P P ( d Q ) Q ρ Q Q d P P Q P Q ( ). Very simiar to casswork probems - () n 0 80 Ω. Given : V and r Ω. V and r Ω 0 Ω Current through eterna resistance (i +i ) and votage drop? Current ectricity

2 .. 6 MHSH TUTOIS SCINC Formua : Σ i 0 & Σi Σ Soution :. Given : 0 V D Current ectricity C B F i i i V V ppying KV to cosed oop B + i r + (i +i ) 0 + i () + (i +i ) 0 0 i + 0i + 0i i + 0 i (i) ppying KV to cosed oop B, + i r + (i + i ) 0. + i () + (i +i ) 0 0 0i + i. (ii) dding (i) and (ii), we get, i + i. i + i. i + i mp 0 6 ppying oop theorem to BCD i (i + i ) Ω Ω 0Ω m 6 Votage drop across each resistance (i + i ) vot i i + i i B V 6. Given : Soution : - (i + i ) - i + - i 0-00 (i + i ) - i (00) i (00) 0-00i - 00i -0 i + i (i) ppying oop theorem to BF, - (i + i ) - i + - i 0-00 (i + i ) - i (00) + - i (00) 0-00i - 00i - i + i (ii) dding (i) and (ii) i + i 0. i + i i + i V r Ω 0Ω Ω I, I 0, I? I I B I et I and I be the currents through resistors and I be the currrent drawn from ce. t node B, using Kirchoff s current aw, I I I 0 0. i + i 7. m I I + I... (i) r C D F I

3 MHSH TUTOIS SCINC.. 7 ppying Kirchoff s votage aw to oop, BCF, I I r I (I + I ) + 0 (using (i)) I + I... (ii) ppying Kirchoff s votage aw to BDCB, I + I 0 I + 0I 0 I I Using (iii) in (ii), (.)I + I 7.6 I I 7.6 Using (iv) in (iii) I I (iii) (iv) I I + I I I be the current r be the interna resistance be the MF of the ce be eterna resistance ccording to kirchoff s votage aw, I (r + ) 0 In case (i), I 0., Ω I (r + ) 0.(r + )...(i) In case (ii) I 0., Ω I (r + ) 0.(r + )...(ii) from (i) and (ii) 0. (r + ) 0.(r + ) r + r + r Ω Substituting r in (i) 0. ( + ) 8. I g? Soution :. V. V I I Using Kirchoff s Votage aw, In oop BD : I 0I g + I 0 I + I 0I g I + I I g... (i) In oop BCDB : 0(I I g ) + 0 (I + I g ) + 0I g 0 0I + 0I g + 0I + 0I g + 0I g 0 I I I g... (ii) dding (i) and (ii), we get, I 6I g...(iii) From (ii), I I + I g Using (iii), I (6I g ) + (I g ) 6I g...(iv) I + I...(given) 6I g + 6I g...using (iii) and (iv) I g I g I g B I I g Ω 0 Ω Ω D I g G G0Ω I +I g 0 Ω C Current ectricity

4 .. 8 MHSH TUTOIS SCINC 9. ppying KV in oop BC ; I (I I g ) 0 6I I g... (i) ppying KV in oop DC ; (I I ) (I I + I ) g 0 I + I + I g... (ii) ppying KV in oop BD ; I 0I g + (I I ) 0 I I + 0I g 0...(iii) Soving equations (i), (ii) and (iii) simutaneousy, we get ; I I I g Current ectricity I Ω I I I Ω V B G D I g Ω Ω I I +I g I I g C 0. Soution : F I/6 G I/6 I/ I/6 I/ I/ B I/ I/6 I I/6 H I/ D I/6 C I/ I et be the resistance of each conductor 6Ω and be the resistance across H i.e. I between two diagonay opposite corners of cube. et I be current suppied by the ce which gets divided in three equa parts I each aong B, D and F as shown in Fig. ppying Kirchhoff s second aw to the mesh DCH I I I I 0 6 I I ( + + ) H Ω. Given : et S be shunt across DC to baance bridge B D When bridge is baanced, B BC D DC S S + S Ω Ω ( + S) S S 8 + S 8S 8 S 6 Ω. Given : ρ 0Ω Q Ω S Ω Ω C

5 MHSH TUTOIS SCINC.. 9 S 0Ω Ω? Soution : et resistance connected in parae with S to baance the network S esistance of parae S and S + for wheatstone s Baanced n/w P Q 0 S S S S + S S Ω. Given : 0 Ω, 0 Ω, 0 Ω and Ω form Wheatstone s newtork. esistance to be connected in branch DC to baance network? Formua : Baancing condition for Wheatstone s bridge Soution : et S be shunt connected across Ω to baance bridge 7. Given : et X be unknown resistance in right gap of metre bridge. When bridge is baanced then, B BC 0 X X X 7 X D DC S S 0 S 0 + S S S S 0 S 0 Ω 0 7 X 70Ω. et G be unknown resistance in eft gap and 8Ω in right gap of metre bridge. Nu point is obtained cm, from eft end of wire. G cm ; 00 G 8Ω In Kevin s method when bridge is baanced, G G G G G 00 - G Current ectricity

6 .. 60 MHSH TUTOIS SCINC G Current ectricity 990 G Ω 6. Given : i) qua engths of manganin & nichrome are connected in eft and right gaps of metre bridge. ii) Nu point is obtained 0cm from eft end of bridge wire. 0 cm iii) ρ m 0-8 Ωm ρ n 0-6 Ωm d m d n? Formua : i) ρ & π d ii) Baancing condition for metre bridge. Soution : et m and n be resistances of manganin & nichrome wires connected in two gaps of metre bridge m n 00- ρ m. m m 0 ρ n. n 00 0 n ρ m. m πd /. n ρ n n πd / m ρ m. m. ( d n ) ρ n n d m... m n, so ρ m 0-8 Ωm ρ n 0-6 Ωm 0-8 m ( d n ) 0-6 m d m ( d n ) 0-6 d m 0-8 ( d m ) 8. Given : Ω 0cm 0-8 d n 0-6 ( d m ) 0 d n 00 0 d m d n 0 7. Given : et be resistance of given wire, 6Ω. If the wire is bent into circe, resistance between diametricay opposite points woud be. 8 8 Ω If the circuar oop with diametricay opposite points is connected in eft gap and unknown resistance X in right gap to obtain nu point at midde of wire. ( ) ( ) X 00-0 (... d 0cm) X 0 X Ω r 00 60cm

7 MHSH TUTOIS SCINC.. 6 Soution : et be the resistance of wire. When the wire is bent in the form of ring and connected in eft gap with diametricay opposite points between and B The resistance is equivaent to two resistances each of haf the resistance in parae so the resistane in eft gap is given by, p p p + + Now from baancing condition of meter bridge p p X X 0 X 0 Ω B 9. Soution : et and be the resistance of two coi when they are in series in one gap, the resistance becomes s + The nu point is at the centre i.e. 0cm from either end 00Ω X s s + 00Ω...(i) 00 When two cois are connected in parae in one gap, the resistance beocmes, p. +. The resistance in other gap is changed by 8Ω i.e 00 ± 8 But p is aways ess than s < Ω P s...from (i) (00 ) ( 80) ( 0) 0 80Ω 0Ω or 0Ω or 80Ω Thus resistance are 80Ω and 0Ω Current ectricity s

8 .. 6 MHSH TUTOIS SCINC 0. Soution : Case () : is connected in eft gap in ight gap 70cm cm Current ectricity (i) 7 Case () : + 0cm ( ) ( + ) Ω 7 7. Ω. Ω,. Ω 7...(ii). Given : y X + 0 Y + 0 X? Y? Formua : 6 X X Y Y Soution : From st condition X Y X Y... (i) From nd condition X + 0 Y (X + 0) (Y + 0) 6 6X + 80 Y + 0 (X) + 80 Y + 0 (Y) + 80 Y + 0 Y + 80 Y + 0 Y 0 Ω... (ii) From equation (i) and (ii) X (0) X 60 X 0 Ω. Given : Formua : 0 Ω X 0 cm 60 cm esistance of entire wire? X. X

9 MHSH TUTOIS SCINC.. 6 Soution : et, ength of smaer piece of wire ength of arger piece of wire Correspondingy resistance of the pieces is and. The wires are connected paraey in the eft gap of meter bridge X X X. Ω In the eft gap, equivaent resistance is given by X Ω. Given : Now entire resistance of wire Ω 0 Ω X 0 cm 60 cm i) Unknown resistance (X)? ii) Shift in the position of the nu point a) when the resistances in both the gaps are increased by Ω and b) when the resistance in each gap is shunted by a resistance of 8 Ω. Formua : Soution : X. X i) X. X X X 0 Ω ii) a) When the resistance in both the gaps are increased by Ω X X Ω Ω Since X + 00 X 00 X. X cm X X.7 cm Shift in nu point.7 0 Shift in nu point.7 cm towards right b) When the resistance in each gap is shunted by a resistance of 8 Ω new resistance in eft gap X Current ectricity

10 .. 6 MHSH TUTOIS SCINC X X.7 Ω New resistance in right gap Ω X X When 0Ω is shunted by 0Ω, the resistance in eft gap becomes 0Ω X P 0 0 X p 0Ω end. et be the baancing ength from eft X P cm from eft end. Shift nu point is 0 cm towards eft X cm X 7. cm shift in nu point 7. 0 shift in nu point 7. cm towards right. Soution : X 0Ω, 0Ω et be baancing ength from eft end of wire. X cm. from eft end.. Given : d d? Formua : ρ Soution : ρ y ρ. πr ρ π d Current ectricity

11 MHSH TUTOIS SCINC.. 6 ρ π d d...(since resistance wire of same materia, ρ is same) d d...(where and are the resistances connected in eft and right gap) d d Now, By the baanced condition od Wheatstone bridge, y y y...(i) Now, + y...(since its meter bridge) +...From (i) 0. m 6. Given : 0m V B 6V 80 cm.8 m i) K V B ii) K Soution : i) K K V B 6V 0m K 0.6 V/m K 6 0 V/cm 0 K V/cm ii) K V 7. Given : 0Ω 0m V K µv/mm K 0-6 V 0 - m K 0 - V/m h? Formua : i) σ ii) i iii) K i.σ Soution : σ + h Current ectricity

12 .. 66 MHSH TUTOIS SCINC σ Current ectricity 0Ω 0m σ Ω/m ii) K i.σ 0 - i. () i 0 - so, i h + h 0 + h 0.00 h 990 Ω 8. Given : V r Ω m Ω (i) Potentia gradient (K)? Formuae : (i) i (ii) σ (iii) K i.σ + r Soution : (i) Current fowing through potentiometer wire, i + r i + ii) so, resistance per unit ength of potentiometer wire is σ σ 7 Ω/ m iii) Potentia gradient aong the wire, V B K K i. i. σ 7 V / m K 0.9 V/m K 0.9 V 00 cm V /cm 9. Given : 0 Ω Formua : 0 m V r Ω. V. V Baancing ength ( ) for sum of & Baancing ength ( ) for difference of & i) σ ii) i + r iii) K i σ iv) + K v) - K Soution : i) σ σ 0 Ω 0 m σ Ω / m

13 MHSH TUTOIS SCINC.. 67 ii) i i + r 0 + r 0 ( 0 - ) 90 r 0 ( - ) i iii) K i. σ K K 0. V / m K 0. V/m iv) + K. +.. m v) - K m 0. Given : 0 cm 0 Ω 90 cm Interna resistance of ce (r)? Formua : r ( - ) Soution : The interna resistance of ce is given by, r ( - ) r 0 r 6.67 Ω. Given : - % of 0.7 Formua : Soution : Ω r? r ( - ) The interna resistance of a ce is given by, r ( - ) r ( - ) r r Ω. Given : The individua ce method,.8 V 00 cm m 9 V?? Current ectricity

14 .. 68 MHSH TUTOIS SCINC Formua : Soution :. Given : Soution : 0cm 0 Ω Ω cm et be the baancing ength, when the ce is in open circuit r be the interna resistance of ce r ( ) ( 0) ( 00) In individua ce method, m. cm. m r r. Ω Given : V r Ω m Ω (i) Potentia gradient (K)? Formuae : (i) i (ii) σ (iii) K i.σ + r Soution : (i) current fowing through potentiometer wire, i + r i + ii) so, resistance per unit ength of potentiometer wire is σ σ 7 Ω/ m iii) Potentia gradient aong the wire, K K V B i. i. σ 7 V / m Current ectricity

15 MHSH TUTOIS SCINC.. 69 K 0.9 V/m K 0.9 V 00 cm V /cm. Given : σ Ω /m. V 80 m.8 m.08 V Current through the wire (i)? Baancing ength for ce of e.m.f. (.08 V)? Formuae : i) K ii) K i. σ iii) K Soution : K. K (.8) K..8 V/m... (i) so, K i. σ i. σ... from (i) i... (... σ ) 0. Formua : K K V B Soution : i) Potentia gradient aong the wire, V B K K 6 V 0 m 0.6 V / m ii).m.f. of a ce, K V so, K.08.6 m 6 cm 6. Given : 0 m V B 6 V 80 cm.8 m.m.f. of a ce ( )? Current ectricity