Question

Three resistors 5 ohm 10 ohm 15 ohm are connect in series,potential drop across the 10 ohm resistor is 4 volt find the current in the circuit and potential drop across the 5 ohm resistor.​

Answer 1

Answer:

Current in the circuit is 0.4 A .

Potential drop across 5 Ω resistor is 2 V.

Explanation:

To find :

• Current in the circuit.
• Potential drop across the 5 Ω resistor.

Given that,

There are three resistors of 5 Ω, 10 Ω and 15 Ω.

Potential drop across the 10 Ω resistor is 4 V.

We know,

Same current flows through all the resistors in series circuit.

We know,

Ohms law,

• V = IR or, I = V/R

[Where, V is voltage, I is current and R is resistance].

Current flowing through 10 Ω resistor :

→ I = 4 V/10 Ω

→ I = 2/5

→ I = 0.4 A

∴ Current in the circuit is 0.4 A .

Now,

Potential drop across the 5 Ω resistor :

→ V = 0.4 A × 5 Ω

→ V = 2 V

∴ Potential drop across 5 Ω resistor is 2 V.

Answer 2

Answer:

Given :-

• Three resistors 5 ohm, 10 ohm, 15 ohm are connect in series.
• Potential drop across the 10 ohm resistor is 4 volt.

To Find :-

• What is the current in the circuit and potential drop across the 5 ohm resistor.

Solution :-

$\clubsuit$ Voltage Formula :

$\mapsto \sf\boxed{\bold{\pink{V =\: IR}}}$

where,

• V = Voltage (v)
• I = Current (A)
• R = Resistance (Ω)

Solution :-

First, we have to find the current in the circuit :

Given :

• Voltage (v) = 4 Volt
• Resistance (R) = 10 Ω

According to the question by using the formula we get,

$\longrightarrow \bf V =\: IR$

$\bigstar\: \: \sf\boxed{\bold{\pink{I =\: \dfrac{V}{R}}}}$

$\longrightarrow \sf I =\: \dfrac{4}{10}$

$\longrightarrow \sf\bold{\red{I =\: 0.4\: A}}$

${\small{\bold{\underline{\therefore\: The\: current\: in\: the\: circuit\: is\: 04\: A\: .}}}}$

Now, we have to find the potential drop across the 5 ohm resistor :

Given :

• Current (I) = 0.4 A
• Resistance (R) = 5 Ω

According to the question by using the formula we get,

$\longrightarrow \sf V =\: 0.4 \times 5$

$\longrightarrow \sf V =\: \dfrac{4}{10} \times 5$

$\longrightarrow \sf V =\: \dfrac{4 \times 5}{10}$

$\longrightarrow \sf V =\: \dfrac{2\cancel{0}}{1\cancel{0}}$

$\longrightarrow \sf V =\: \dfrac{2}{1}$

$\longrightarrow \sf\bold{\red{V =\: 2\: V}}$

${\small{\bold{\underline{\therefore\: The\: potential\: drop\: across\: the\: 5\: ohm\: resistor\: is\: 2\: V\: .}}}}$