Question

draw a schematic diagram of a circute consisting of a batterybof 3 cells of 2v each ,a combination of three resistors of 10 ohm 20 ohm 30 ohm connected in parallel,a plug key and an ammeter all connected in series .............find thevalues of current through each resistor, total current in the circute ,total effective resistance of the circuit​

Answer 1

Answer:

Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, an ammeter and a plug key; all connected in series.

Draw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 Ω resistor, a 10 Ω resistor and a 15 Ω resistor, an ammeter and a plug key; all connected in series. Also connect a voltmeter to record the potential difference across the 15 Ω resistor and calculate :

(i) the electric current passing through the above circuit and

(ii) potential difference across 50 resistor when the key is closed.

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1 Answer

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answered Nov 10, 2017 by priya12 (-12,637 points)

10 V Battery ; Rest Components

Equivalent resistance = R1 +R2 +R3

= 5 + 10 + 15

= 30 Ω

Current in the circuit , I = V / R

(ii) Potential Differences across 5Ω resistor , V = IR

Answer 2

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

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

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

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:

$V_B=3\times V_c$

$V_B=3\times 2$

$V_B=6V$

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:

$i=i_1+i_2+i_3$

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

$i=\frac{11}{10}\ A=1.1\ A$

Now the effective resistance of the circuit:

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

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

$R=\frac{60}{11}=5.45\Omega$

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