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An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 respectively, then the ratio of the currents passing through the wires will be

An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 respectively, then the ratio of the currents passing through the wires will be

Byju's Answer Standard XII Physics

Partially Filled Dielectrics

An electric c... Question

An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wires are in the ratio of

4 / 3 and 2 / 3

respectively, then the ratio of the currents passing through the wires will be -

A 1 / 3 B 1 / 5 C 8 / 9 D 2 Open in App Solution

The correct option is A

1 / 3

Since wires are connected in parallel, the potential difference across the wires will be the same.

I 1 = V R 1 and I 2 = V R 2 ⇒ I 1 I 2 = R 2 R 1 Here, R 1 = ρ l 1 A 1 = ρ l 1 π r 2 1 , and R 2 = ρ l 2 π r 2 2 ⇒ I 1 I 2 = ρ l 2 π r 2 2 ρ l 1 π r 2 1 = l 2 r 2 1 l 1 r 2 2 ⇒ I 1 I 2 = 3 4 × ( 2 3 ) 2 = 1 3 Hence, option ( A )

is the correct answer.

Suggest Corrections 3

SIMILAR QUESTIONS

Q. An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wires are in the ratio of

4 / 3 and 2 / 3

respectively, then the ratio of the currents passing through the wires will be -

Q. The same mass of copper is drawn into wires

2 m m and 3 m m

thick. The wires are connected in series and current is passed through them. The ratio of heats produced in two wires is

Q. Two wires of same material having lengths and radii in the ratio of

3 : 4 and 3 : 2

respectively are connected in parallel with a potential source of

6 V

. The ratio of currents flowing through them,

I 1 : I 2 = _________.

स्रोत : byjus.com

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 , then the ratio of the currents passing through the wire will be

Click here👆to get an answer to your question ✍️ An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3 , then the ratio of the currents passing through the wire will be

Question

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be

A

3

B

3 1 ​

C

9 8 ​

D

2

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is B)

R=ρ

A l ​ ⟹R=ρ πr 2 l ​ ⟹R∝ r 2 l ​ Hence, R R ′ ​ = r ′2 l ′ ​ l r 2 ​ R R ′ ​ = l l ′ ​ (r ′ ) 2 r 2 ​ ⟹ R R ′ ​ = (2/3) 2 4/3 ​ ⟹ R R ′ ​ =3 Since, I∝ R 1 ​ Hence, I I ′ ​ = 3 1 ​ Answer-(B)

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An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If lengths and radii of the wires are in the ratio of 4 : 3 and 2 : 3 , then ratio of the currents passing through the wires will be

we known that R prop (l)/(r^(2)) rArr (R(1))/(R(2))=(l(1))/(l(2))xx(r(2)^(2))/(r(1)^(2))=(4)/(2)xx(9)/(4)=(3)/(1) Since it is parallel current, inversely proportional to resistance V=IR R prop (1)/(i), (R(1))/(R(2))=(i(2))/(i(1))rArr (3)/(1) = (i(2))/(i(1))rArr (i(1))/(i(2))=(1)/(3).

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An electric current is passe...

An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If lengths and radii of the wires are in the ratio of 4 : 3 and 2 : 3 , then ratio of the currents passing through the wires will be

Updated On: 27-06-2022

( 00 : 09 ) ADVERTISEMENT Text Solution Open Answer in App Solution we known that R∝ l r 2 ⇒ R 1 R 2 = l 1 l 2 × r 2 2 r 2 1 = 4 2 × 9 4 = 3 1

R∝lr2⇒R1R2=l1l2×r22r12=42×94=31

Since it is parallel current, inversely proportional to resistance

V=IR V=IR R∝ 1 i , R 1 R 2 = i 2 i 1 ⇒ 3 1 = i 2 i 1 ⇒ i 1 i 2 = 1 3

R∝1i,R1R2=i2i1⇒31=i2i1⇒i1i2=13

.

Answer

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