Two resistors 400ohm and 800ohm nbsp-connected in series with a battery of 6v- an ammeter of 10ohm is used to to measure the current what will the ammeter nbsp-read- a voltmeter of 10000ohm nbsp-is used to find potential difference across 400ohm- what will it read

Dear Student Total resistance of circuit is 400 +800 +10 ohm =1210 ohm nbsp; (As resistors are connected in series with ammeter) Current through circuit or ...

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Standard X

Physics

Ohm's Law

two resistors...

Question

two resistors 400ohm and 800ohm connected in series with a battery of 6v. an ammeter of 10ohm is used to to measure the current what will the ammeter read? a voltmeter of 10000ohm is used to find potential difference across 400ohm. what will it read

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Solution

Dear Student

Total resistance of circuit is 400 +800 +10 ohm =1210 ohm (As resistors are connected in series with ammeter)

Current through circuit or each resistor or ammeter will be I = V/R = 0.0049 A = 4.9 x 10^-3 A

Now when voltmeter is connected across 400 ohm resistance.
Then total resistance of voltmeter and 400 ohm resistor is

$\frac{1}{R} = \frac{1}{400} + \frac{1}{10000} = \frac{10400}{400 \times 1000} R = 384 . 6 o h m C u r r e n t I t h r o u g h c i r c u i t = \frac{6}{384 . 6 + 400} = 0.0051 A N o w l e t c u r r e n t I_{1} p a s s e s t h r o u g h v o l t m e t e r a n d I_{2} p a s s e s t h r o u g h 400 o h m r e s i s t o r . V o l t a g e a c r o s s b o t h w i l l b e s a m e . S o 400 I_{2} = 10000 I_{1} I_{1} = 0.04 I_{2} B u t I = I_{1} + I_{2} I_{2} + 0.04 I_{2} = 0.0051 I_{2} = \frac{0.0051}{1.04} = 0.0049 T H e r e f o r e p o t e n t i a l d i f f e r e c n e a c r o s s 400 o h m = 400 I_{2} = 400 \times 0.0049 = 1.95 V T h e r e f o r e r e a d i n g o f a m m e t e r = 1.95 V o l t$
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