Question

Calculate the current flowing through 4 ohm resistor:

Answer 1

The figure given in the question above contains resistors , a voltage source and ammeters. From here we can conclude that when ammeters are connected in the circuit and no voltmeter is present ,it means the entire circuit is in series connection because ammeters are connected in series connection only.
In series connection , the magnitude of current remains the same .Therefore ,we only have to apply KVL(Kirchhoff’s Voltage Law) to find out the current in the 4ohm resistor.

Complete step by step solution :
We have $8V$ battery, $4ohm$ resistor connected in series with the voltage source , two $8ohm$ resistors connected in parallel with each other and the combination of both in series with the entire circuit.

First of all we will solve the parallel combination of $8ohm$ resistors into a single resistor.
$\dfrac{1}{8} + \dfrac{1}{8} = 4$ (in parallel connection ,we use the fractions of the given magnitude to calculate the equivalent resistor)

Now let's apply the KVL in the circuit , which states that:
“Sum of all the voltage drops in the circuit is equal to zero “.
Keep the unknown current flowing in the circuit as $i$ .

Sign convention are as follows :
While moving in the clockwise direction ,if we come across a positive sign first which drops to negative sign we use negative sign in the equation . If we come across negative signs followed by positive then we will use positive signs in the equation .

$8 - 4i - 4i = 0 \\\ \Rightarrow 8 - 8i = 0 \\\ \Rightarrow i = \frac{8}{8} = 1 \\\ \Rightarrow i = 1A \\\$

Note:
Ammeter measures the magnitude of current flowing in the circuit and always connected in series in the circuit . Voltmeter measures the magnitude of voltage drop across the connected circuit and is always connected in parallel in the circuit. When resistors are connected in series , the given magnitudes are added directly without taking reciprocal value to find an equivalent resistor in the circuit.