Question

A battery of 9 V is connected in series with resistors of 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω and 12
Ω respectively. How much current would flow through the 12 Ω resistor?

Answer 1

Answer:

Explanation:

ANSWER:

Voltage - 9 volts

Resistances - 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω and 12

Ω respectively.

Current through 12 Ω resistor  = ?

To find this we need to remember a formula that is:-

$\boxed{\texttt{Voltage = Current * Resistance }}$ or in symbolic form,

$\boxed{\texttt{V = IR}}$

Total resistance in series = 0.2+0.3+0.4+0.5+12Ω

Total Equivalent Resistance = $\boxed{\texttt{13.4 ohm}}$

Using the above formula, Placing the given values,

9 = 13.4 * Current

Current = $\frac{9}{13.4} = \boxed{\texttt{0.67 Amperes}}$ is the required current.

As all resistances are connected in series combination,

$\boxed{\textsf{Current through each resistors will be same}}$, and hence current flowing through 12 ohm resistor will be 0.67 Amperes.

Things to Note:-

• In series combination current flowing through each resistances will be same
• In parallel according to resistance the current flowing through each resistance will be different.
• In series combination simple addition is done to find the equivalent resistance
• In parallel combination we reciprocal the values to find the equivalent resistance
• Voltage = Current*Resistance

Answer 2

Answer:

Explanation:

Solution :-

Here,

We have,

Potential Difference = 9 V

R₁ = 0.2 Ω

R₂ = 0.3 Ω

R₂ = 0.4 Ω

R₄ = 0.5 Ω

R₅ = 12 Ω

Total Resistance = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now,

According to the Ohm's Law

We know that,

I = V/R

⇒ I = 9V

⇒ I = 9V/13.4 Ω

⇒ I = 0.67 A.

Hence,  current flow through the 12 Ω resistor is  0.67 A.