# Electrical Engineering - Ohm's Law - Discussion

### Discussion :: Ohm's Law - General Questions (Q.No.12)

12.

You are measuring the current in a circuit that is operated on an 18 V battery. The ammeter reads 40 mA. Later you notice the current has dropped to 20 mA. How much has the voltage changed?

 [A]. 9 V [B]. 900 mV [C]. 0 V [D]. 18 V

Explanation:

No answer description available for this question.

 Nikul Prajapati said: (May 24, 2011) 40mA - 18V then 20mA - (?) 20*18/40 = 9V

 Sunand Behera said: (Aug 30, 2011) Circuit resistance remains constant and then follow as Nikul's answer.

 Praveen2Kanna said: (Sep 10, 2011) V=I*R;==> R=V/I; V=18V INITIALLY CURRENT IS 40mA SO RESISTANCE = 18/(40*10^-3)=450 ohm THEN CURRENT DROPS TO 20mA AS RESISTANCE REMAIN SAME 450 ohm V=(20*10^-3)*450 = 9V

 Eeebhola said: (Oct 31, 2011) We know that from ohms law V=IR or we can say V directly proportional to current. Thats why answer (a).

 Kavitha said: (Mar 28, 2012) V1/I1=V2/I2 V2=(V1*I2)/I1 V2=(18*20)/40=9V

 Ashok said: (Apr 7, 2012) V=IR(OHM'S LAW) R=V/I TAKE RATIO:V1/I1=V2/I2 V2=(V1*I2)/I1 V2=(18*20)/40 V2=9V.

 Mathan Senthil said: (Apr 8, 2012) (V1/I1)=(V2/I2) THEREFORE V2=(V1*I2)/I1 V2=(18*20*10^-3)/(40*10^-3)= 9V

 Sunil Pant said: (Apr 17, 2012) 40mA -> 18V Then 20mA -> (?) 20*18/40 = 9V

 Ashish said: (May 10, 2012) Here it is not clear that weather variable or constant load is connected. Say earlier load which takes 40ma is reduced i.e. load is cut, but voltage is constant, so my view we can also say that voltage are remain same.

 Darshanmaths said: (Aug 13, 2012) V1 = 18V I1 = 40mA. V2 = ? I2 = 20mA. As per ohm's law, As resistance remains constant. V1/I1 = V2/I2. Hence substituting, 18/40 = V2/20. Hence, V2=0.45*20 = 9V. Hence V2 = 9V.

 Julea said: (Mar 10, 2015) How did they get 9v?

 Pravi said: (May 7, 2015) First find the resistance, R = 18/(40*10^-3) = 450. V' = 450*20*10^-3. Then, V' = 0.45*20. V' = 9 v.

 Raji said: (Aug 30, 2015) Current changed but the resistance remains constant. Hence find resistance from given data i.e. v=18V, i=40mA using the formula V=IR. Now find voltage for new current (20mA) flowing through same resistor using same formula.

 Hardik said: (Oct 5, 2015) V = 18 v, I = 40 mA, R = ? V = IR. R = V/I. = 18/40*10^-3. = (18/40)*10^3. = 18000/40. = 450 ohm. Then, for V=?, R= 450, I=20 mA. V = IR. = (20*10^-3)*450. = 9000*10^-3. v = 9V.

 Krishna said: (Feb 23, 2016) Or you can say V is directly proportional to I. So when the current drops so as the voltage.

 Vinodroyal said: (Feb 25, 2016) If current increase voltage decreases.

 Roderick Acuesta said: (Mar 5, 2016) The problem is not clear because not tell if the load is variable.

 S K S said: (Apr 11, 2016) Voltage change 18 - 9 = 9 volt.

 Estrellanes said: (May 3, 2016) The half of 40 is 20, and the half of 18 is 9. So the answer is A) 9v.

 Roshan said: (Jun 7, 2016) Becuase ohms law online define a material body, not a practical area. So if voltage increasing it not necessary to increases current and vice versa.

 Jagan said: (Nov 20, 2016) Thanks for all the answer and explanation.

 Srk said: (May 24, 2017) v1=18 i1=40ma,i2=20ma,v2=? R=18/40ma = 450. V2 = i2 * r = 20m * 450 = 9a.

 Yaswanth Pagadala said: (Sep 24, 2017) Thanks for all.

 Swetha said: (Aug 10, 2018) Vg is directly proportional to I. Ohm's law applies. V1/V2=I1/I2 substitute this formula....

 Ladli said: (Aug 21, 2018) Here, the load is constant(R). So,v/i=R=v'/i'. 18/40mA=v'/20mA. Hence, v'=9volt.

 Hardik said: (Mar 10, 2019) V = IR. V directly proportional to I. So, Apply proportional rules.

 Azeez said: (Aug 4, 2019) V=IR. i.e. V is directly proportional to I. Current is changing from 40 mA to 20 mA. V1/V2= I1/I2. 18/V2=40mA/20mA. V2=9V.