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Two resistors 400 ohm and 800 ohm are connected in series with a 6V battery. It is desired to measure the current in the circuit. An ammeter of 10 ohm resistance is used for this purpose. What will be the reading in the ammeter? Similarly if a voltmeter of 10,000 ohm resistance is used to measure the potential difference across 400 ohm, what will be the reading of the voltmeter ?

```
6 years ago

``` Regards,Nirmal SInghAskiitians Faculty
```
6 years ago
```							Dear StudentTotal 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 ANow when voltmeter is connected across 400 ohm resistance.Then total resistance of voltmeter and 400 ohm resistor isR = 384. 6 ohmCurrent I through circuit = = 0.0051 A 6384/ 6+400Now let current passes through voltmeter and passes through 400 ohm resistor. I1 I2 V oltage across bothwill be same.So400 I2 = 10000I1I1 = 0.04 I2ButI = I1 +I2I2 +0.04 I2 = 0.0051I2 = = 0.00490.00511.04THerefore potential differecne across 400 ohm = 400 I2 = 400×0.0049 = 1.95 VTherefore reading of ammeter = 1.95 V olt
```
3 years ago
```							Dear Student,Please find below the solution to your problem.Given,R1 ​= 400Ω, R2 ​= 800ΩE = 6V1stCaser = 10Ωi = 6/(10 + 400 + 800) ​= 6/(1210) ​= 4.96mA2ndCaser = 10000ΩReq​ = (10000×400)/10400​ + 800 = 30800​/26i = 6×26/30800​i 400Ω ​= (25/26​)×(6×26/30800)​V 400Ω ​= i×400 = 150​×400/30800 = 150/77 ​= 1.95VHence option B is the correct answer.Thanks and Regards
```
2 months ago
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