In the circuit shown below, calculate

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Question

$(a)$ In arm $CE$ resistors of value $2$ and $4\, ohm$ are in series, therefore,

Total resistance in $\operatorname{arm} CE =2+4=6 ohm$

$(b)$ To

Vedclass · Accepted Answer

Vedclass · Answer

$(a)$ In arm $CE$ resistors of value $2$ and $4\, ohm$ are in series, therefore,

Total resistance in $\operatorname{arm} CE =2+4=6 ohm$

$(b)$ Total resistance of the circuit is

$R _{ P }=\frac{6 	imes 6}{6+6}=3 \Omega$

Therefore, total current in the circuit is

$I=\frac{V}{R}=\frac{3}{3}=1 A$

$(c)$ Since the arms $CE$ and $AB$ have the same resistance, therefore, this current is split equally amongst the two. Hence, current in arms $AB$ and $CE$ is $0.5 A$

In the circuit shown below, calculate

$(a)$ total resistance in arm $CE,$

$(b)$ total current drawn from the battery, and

$(c)$ current in each arm, i.e., $A B$ and $C E$ of the circuit.

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