Question

please solve this question.
chapter: electricity class:10

Answer 1

Answer:

14 wount of 0.75 mbiu is design ya filament of a eleduc tub fpec 5 minuty Find out the amount eluding charge that flows, the clatronic sinciitt

Answer 2

Question

Five resistors of different resistance are connected together as shown in figure. A 12V battery is connected to the arrangement Calculate

(a)the total resistance in the circuit,

(b) the total current flowing in the circuit.

Answer

Given :

In the given figure,

• R₁ = 10Ω
• R₂ = 40Ω
• R₃ = 30Ω
• R₄ = 20Ω
• R₅ = 60Ω

voltage of the battery, V = 12V.

To find :

(a) Total Resistance in the circuit

(b) Total current flowing in the circuit

Solution :

(a) In the given figure there are 2 setup of connection. In one setup R₁ and R₂ are connected in parallel combination.

» The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

$\dashrightarrow \sf \dfrac{1}{R _{12}} =\dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\dashrightarrow \sf \dfrac{1}{R_{12}} =\dfrac{1}{10} + \dfrac{1}{40}$

$\dashrightarrow \sf \dfrac{1}{R_{12}} =\dfrac{4 + 1}{40}$

$\dashrightarrow \sf \dfrac{1}{R_{12}} =\dfrac{5}{40}$

$\dashrightarrow \sf \dfrac{1}{R_{12}} =\dfrac{1}{8}$

$\dashrightarrow \sf R_{12} =8 \: ohms$

similarly,

In another setup,

there are three resistors R₃, R₄ and R₅ are connected in parallel combination,

$\dashrightarrow \sf \dfrac{1}{R_{345}} =\dfrac{1}{R_3}+ \dfrac{1}{R_4} + \dfrac{1}{R_5}$

$\dashrightarrow \sf \dfrac{1}{R_{345}} =\dfrac{1}{30}+ \dfrac{1}{20} + \dfrac{1}{60}$

$\dashrightarrow \sf \dfrac{1}{R_{345}} =\dfrac{2 + 3 + 1}{60}$

$\dashrightarrow \sf \dfrac{1}{R_{345}} =\dfrac{6}{60}$

$\dashrightarrow \sf \dfrac{1}{R_{345}} =\dfrac{ 1}{10}$

$\dashrightarrow \sf R_{345} =10 \: ohms$

Now both set R₁₂ and R₃₄₅ come in series combination, so total resistance in the circuit,

$\dashrightarrow\sf R_{eq} = R_{12} +R_{345}$

$\dashrightarrow\sf R_{eq} = 8 + 10$

$\dashrightarrow\sf R_{eq} = 18 \: ohms$

thus, total resistance of the circuit is 18 Ω.

(b) According to ohms law,

The potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it provided temperature remains the same .i.e.,

V = IR

here,

• V denotes potential difference
• I denotes current
• R denotes resistance

by substituting all the given values in the formula,

$\dashrightarrow \sf V = IR$

$\dashrightarrow \sf I = \dfrac{V}{R}$

$\dashrightarrow \sf I = \dfrac{12}{18}$

$\dashrightarrow \sf I = 0.67A$

thus, Total current flowing in the circuit is 0.67A.

Remember !

SI unit of Resistance is ohms (Ω)

SI unit of current is Ampere (A)