Question

Resistors R1,R2, and R3 having values 5ohm, 10ohm and 30ohm respectively are connected
in parallel across a battery of 12 volt. Calculate (a) the current through each resistor
(b) the total current in the circuit and (c) the total circuit resistance.​

Answer 1

Given :

▪ Three resistors of resistance 5Ω, 10Ω, and 30Ω are connected in parallel across a battery of 12V.

To Find :

▪ The current through each resistor.

▪ Total current flow in circuit.

▪ Equivalent resistance of circuit.

Solution :

✒ First we have to find out equivalent resistance of circuit after that we can calculate current flow in circuit and in each resistor with the help of ohm's law.

✒ As per ohm's law, current flow in resistor is directly proportional to the potential difference. (at constant temperature)

✴ Equivalent resistance (Parallel) :

☞ 1/Req = 1/R1 + 1/R2 +...+ 1/Rn

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▶ Equivalent resistance of circuit :

→ 1/Req = 1/R1 + 1/R2 + 1/R3

→ 1/Req = 1/5 + 1/10 + 1/30

→ 1/Req = (6 + 3 + 1)/30

→ 1/Req = 10/30

→ Req = 3Ω

▶ Net current flow in circuit :

✏ V = I × Req

✏ 12 = I × 3

✏ I = 12/3

✏ I = 4A

[☞ Potential difference remains same across all resistors in parallel connection.]

▶ Current flow in 5Ω resistor :

→ V = I1 × R1

→ 12 = I1 × 5

→ I1 = 2.4A

▶ Current flow in 10Ω resistor :

✏ V = I2 × R2

✏ 12 = I2 × 10

✏ I2 = 1.2A

▶ Current flow in 30Ω resistor :

→ V = I3 × R3

→ 12 = I3 × 30

→ I3 = 0.4A

Answer 2

Aɴꜱᴡᴇʀ

☞ Current through each resistor = 2.4 A, 1.2 A and 0.4 A

☞ Net resistance = 3 $\sf\Omega$

☞ Net current = 4 A

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Gɪᴠᴇɴ

➳ Resistors are

◕ 5$\sf\Omega$

◕ 10$\sf\Omega$

◕ 30$\sf\Omega$

➳ They are connected in parallel

➳ Voltage (V) = 12 V

_________________

Tᴏ ꜰɪɴᴅ

➤ Current (I) through each resistor?

➤ Total current in the circuit

➤ Net resistance

_________________

Sᴛᴇᴘꜱ

❍ First we shall find the current through each resistor and for that we use the Ohm's Law, which states that the voltage across a circuit is directly proportional to the product the current through the circuit and the resistance provide, that is V = IR

So first the current through the 5$</u><u>\</u><u>s</u><u>f</u><u>\</u><u>O</u><u>m</u><u>e</u><u>g</u><u>a</u><u>$ resistor is,

$\leadsto{} \sf{}12 = i \times5 \\ \\ \leadsto{} \sf \frac{12}{5} = i \\ \\ \pink{\leadsto \sf{}i = 2.4 \:A }$

So the current through the 10$</u><u>\</u><u>s</u><u>f</u><u>\</u><u>O</u><u>m</u><u>e</u><u>g</u><u>a</u><u>$ resistor is,

$\leadsto \sf12 = i \times 10 \\ \\ \leadsto \sf \frac{12}{10} = i \\ \\ \pink{ \leadsto{ \sf{}i = 1.2 \: A}}$

So then the current through the 30 Ω resistor is,

$\leadsto \sf{}12 = i \times 30 \\ \\ \leadsto \sf\frac{12}{30} = i \\ \\ \pink{ \leadsto \sf{} i = 0.4 \:A }$

Now the net resistance (for parallel connection) is given by,

$\underline{ \boxed{ \sf \red{ \frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2} .... \frac{1}{R_n} } }}$

Substituting the given values,

$\mapsto \sf \frac{1}{R_{net}} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} \\ \\ \mapsto \sf \frac{1}{R_{net}} = \frac{6 + 3 + 1}{30} \\ \\ \mapsto \sf {R_{net}} = 3 \Omega$

So then the net current is also found with the help of Ohm's Law

➼ V = IR

➼ 12 = I × 3

➼ 12/3 = I

➼ I = 4 A

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