Question

Draw schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 10 ohm, a 20 ohm and a 30 ohm resistor and a plug key, all connected in parallel, plug key and and ammeter, all connected in series. Use this circuit to find the value of the following: (a) Current through each resistor (b) Total current in the circuit (c) Total effective resistance of the circuit

Answer 1

Answer:

Explanation:

V=IR

I=V/R

I= 2/10 or 3/10 or 4/10

Answer 2

$R=\frac{60}{11} =5.45\Omega$ resultant resistance in the circuit

Total current in the circuit $i=\frac{11}{10} =1.1\ A$

$i_3=\frac{1}{5}\ A$            $i_2=\frac{3}{10} \ A$                 $i_1=\frac{3}{5} \ A$

Explanation:

Given that:

Voltage of the cells in series, $V_c=2\ V$

Combination of the resistances in parallel:

$R_1=10\Omega\\R_2=20\Omega\\R_3=30\Omega$

Also a plug key and an ammeter all in series.

Effective voltage of battery with 3 identical cells:

$V_B=3V_c$

$V_B=2\times 3$

$V_B=6\ V$

Now current through the first resistance:

$i_1=\frac{V}{R_1}$

$i_1=\frac{6}{10}$

$i_1=\frac{3}{5} \ A$

Now current through the second resistance:

$i_2=\frac{V}{R_2}$

$i_2=\frac{6}{20}$

$i_2=\frac{3}{10} \ A$

Now current through the third resistance:

$i_3=\frac{V}{R_3}$

$i_3=\frac{6}{30}$

$i_3=\frac{1}{5}\ A$

therefore the total current in the  circuit:

$i=i_1+i_2+i_3$

$i=\frac{3}{5} +\frac{3}{10}+\frac{1}{5}$

$i=1.1\ A$

Now the effective resistance of the circuit:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30}$

$R=5.45\Omega$

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TOPIC: circuit

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