Question

find the current drawn from the battery by the network of four resistors of 5 ohm each shown in the given figure​

Answer 1

According to the circuit : [ See the attachment ]

At first :

Case-1:

Resistors a,b,c are connected in series. Which total resistance is $R_{abc}$

Case-2:

$R_{abc}$ and resistor d are connected in parallel. Which is $R_{eff}$

Calculations :

Case-1:

${ \boxed{ \begin{array}{cc} R_{abc} = a + b + c \\ \\ = 5 + 5 + 5 \\ \\ = 15 \:ohm \\ \\ \pink{ \boxed{\therefore \: R_{abc} = 15 \: ohm}}\end{array}}}$

Case-2:

${ \boxed{ \begin{array}{cc} \frac{1}{R_{eff}} = \frac{1}{R_{abc}} + \frac{1}{d} \\ \\ = \frac{1}{15} + \frac{1}{5} \\ \\ = \frac{4}{15} \\ \\ \blue{ \boxed{\therefore \: R_{eff} = \frac{15}{4} \: ohm }} \end{array}}}$

So total resistance of the circuit is : $R_{eff}$ = 15/4 ohm

Now,

We know that,

$\orange{ \boxed{ \begin{array}{cc} V = IR_{eff} \\ \\ \implies \: I = \frac{V}{R_{eff}} \\ \\ = \frac{2.5}{ \frac{15}{4} } \\ \\ = \frac{2.5 \times 4}{15} \\ \\ = 0.66 \: A \end{array}}}$

Current of the circuit is I = 0.66 A