Question

23-IF THREE RESISTORS 2,4,8 OHM COMBINED IN SERIES BY A 20 VOLT BATTRY THEN FIND (1)AMOUNT OF

CURRENT (2)VOLTAGE ACROSS EACH RESISTOR​

Answer 1

Given :

▪ Three resistors of resistances 2Ω, 4Ω and 8Ω are connected in series with a battery of 20V.

To Find :

▪ Current flow in the circuit.

▪ Voltage across each resistor.

Concept :

↗ We know that, current flow through each resistor in series connection remains same.

↗ As per ohm's law, Current flow through a conductor is directly proportional to the applied potential difference.

✴ Eq. resistance of series :

$\bigstar\:\underline{\boxed{\bf{\color{lawngreen}{R_s=R_1+R_2+R_3+...+R_n}}}}$

Calculation :

❇ Net current flow :

$\implies\sf\:V=I\times R_s\\ \\ \implies\sf\:V=I\times (R_1+R_2+R_3)\\ \\ \implies\sf\:20=I\times (2+4+8)\\ \\ \implies\underline{\boxed{\bf{\color{gold}{I=1.428A}}}}\:\gray{\bigstar}$

❇ Voltage across 2Ω resistor :

$\mapsto\sf\:V_1=I\times R_1\\ \\ \mapsto\sf\:V_1=1.428\times 2\\ \\ \implies\underline{\boxed{\bf{\color{navy}{V_1=2.85V}}}}\:\gray{\bigstar}$

❇ Voltage across 4Ω resistor :

$\longrightarrow\sf\:V_2=I\times R_2\\ \\ \longrightarrow\sf\:V_2=1.428\times 4\\ \\ \longrightarrow\underline{\boxed{\bf{\color{aqua}{V_2=5.71V}}}}\:\gray{\bigstar}$

❇ Voltage across 8Ω resistor :

$\dashrightarrow\sf\:V_3=I\times R_3\\ \\ \dashrightarrow\sf\:V_3=1.428\times 8\\ \\ \dashrightarrow\underline{\boxed{\bf{\color{crimson}{V_3=11.42V}}}}\:\gray{\bigstar}$

Answer 2

Given that ,

" Three resistors 2 , 4 , 8 Ω combined in series by a 20 V battery  "

We need to find the

1 ) Amount of current

If 3 resistors R₁ , R₂ , R₃ , ..... are connected in series ,

Then ,

$\bf R_{eq}=R_1+R_2+R_3+.....$

$\implies \bf R_{eq}=2+4+8\\\\\implies \bf R_{eq}=14\ \Omega$

Now , Use Ohm's law to find amount of current ,

$\implies \bf V=IR_{eq}\\\\\implies20=I*14\\\\\implies \bf{\red{ I=1.43\ Amperes}}$

2 ) Voltage across each resistor

$\underbrace{\bf Voltage\ across\ 2\ \Omega :}\\\\\implies \bf V=IR\\\\\implies V=1.43*2\\\\\implies V=2.86\ V\\\\\underbrace{\bf Voltage\ across\ 4\ \Omega:}\\\\\implies \bf V=1.43*4\\\\\implies \bf V=5.72\ V\\\\\underbrace{\bf Voltage\ across\ 8\ \Omega :}\\\\\implies \bf V=1.43*8\\\\\implies \bf V=11.44\ V$