001 (part 1 of 2) 10.0 points A battery with an emf of 8.3 V and internal resist

Question

001 (part 1 of 2) 10.0 points A battery with an emf of 8.3 V and internal resistance of 0.57 is connected across a load resistor R. If the current in the circuit is 1.35 A, what is the value of R? Answer in units of 002 (part 2 of 2) 10.0 points Wh at power is dissipated in the internal re- sistance of the batterv? Answer in units ofW. 003 10.0 points Consider the circuit 31.7 63.4 What is the equivalent resistance between the points a and b? Answer in units of 004 (part 1 of 2) 10.0 points A lamp having a resistance of 15 is con- nected across a 18 V battery What is the current through the lamp? Answer in units of A 005 (part 2 of 2) 10.0 points What resistance must be connected in series with the lamp to reduce the current to 0.42 A? Answer in units of 006 10.0 points Resistances of 2.1 , 4.8 , and 5.2 and a 24.3 V battery are all in series Find the potential difference across the first (21 ) resistor Answer in units ofV.

Explanation / Answer

will answer 4 parts as per Chegg's policy

a) 8.3 = 1.35 ( R +0.57)

R = 5.578 ohms apprx

b) P = I^2r = 1.35^2 ( 0.57) = 1.04 watts

Q:3 equivalent resistance of parallel resistors= 1/63.4 + 1/63.4+ 1/ (95.1) = 2/ 63.4 +1/ 95.1

!/ R = 1/0.042

R = 23.81

equivalent resisrance= 55.5 ohms apoprx

Q: 4I

I = V/R = 18/15 = 1.2 A

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