Question

three resistors of 5 ohm 10 ohm and 15 ohm are connected in series and potential difference of 15 volts is applied through the combination calculate the current flowing through circuit​

Answer 1

AnsWer :

The current flowing through circuit is 0.5A.

GiveN :

• Three resistors of 5Ω 10Ω and 15Ω are connected in series.
• Potential difference of 15 volts is applied through the combination.

SolutioN :

• Series Combination : When current flow only one direction or path not change, it said to be Series Combination.
• Rs = R1 + R2 + R3 + ...Rn.

• Parallel Combination : When current flow different direction or path, it said to be Parallel Combination.
• 1/Rp = 1/R1 + 1/R2 + 1/R3 ... 1/Rn.

We have,

• 5Ω , 10Ω and 15Ω connected in series Combination.

$\tt \dagger \: \: \: \: \: R_S = R_1 + R_2 + R_3 + ...R_n.$

Where as,

• R1 = 5Ω.
• R2 = 10Ω.
• R3 = 15Ω.

$\tt : \implies R_S = 5 + 10 + 15.$

$\tt : \implies R_S =30 \Omega.$

Now, We know

• Ohm's Law.
• V = IR.
• V ( voltage )
• I ( Current )
• R ( Resistance )

$\tt \dagger \: \: \: \: \: V =IR.$

$\tt : \implies 15 = I \times 30.$

$\tt : \implies I = \dfrac{\cancel{15}}{\cancel{30}}$

$\tt : \implies I = \dfrac{1}{2}$

$\tt : \implies I = 0.5A$

Therefore, the value of the current flowing through circuit is 2 A.

Answer 2

${\underline{\sf{Given}}}$

• Three resistors 5 Ω,10 Ω and 15 Ω are connected in series
• And 15 V is applied across the combination

${\underline{\sf{To\:Find}}}$

The current flowing through circuit.

Before solving the question , Let's know about some Combination of resistors .

• Resistors in series and parallel :

1) Resistors in series :

For n resistors connected in series

the equivalent Resistance $R_{eq}$ is given by

$R_{eq}=R_{1}+R_{2}+R_{3}+...+R_{4}$

2) Resistors in parallel:

For n resistors connected in parallel

the equivalent Resistance $R_{eq}$ is given by

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1} } + \dfrac{1}{R_{2} } + \dfrac{1}{R_{3} } + ...... \dfrac{1}{R_{n} }$

${\underline{\sf{Solution}}}$

Let the given resistors be $\sf\:R_{1},R_{2}\:and\:R_{3}$

$\sf\:\implies\:R_{1}=5$Ω

$\sf\:\implies\:R_{2}=10$Ω

$\sf\:\implies\:R_{1}=15$Ω

Since , All the Resistors $\sf\:R_{1},R_{2}\:and\:R_{3}$ are connected in series combination . The equivalent resistance R between the terminals is , Therefore given by :

$\bf\:R=R_{1}+R_{2}+R_{3}$

$\sf\:\implies\:R=(5+10+15)$Ω

$\sf\:\implies\:R=30$Ω

The current supplied by the battery

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

$\sf\:\implies\:I=\dfrac{15}{30}$

$\sf\:\implies\:I=0.5\:A$

Therefore, the current flowing through circuit is 0.5 A