मराठी

Two Identical Cells of Emf 1.5 V Each Joined in Parallel, Supply Energy to an External Circuit Consisting of Two Resistances of 7 Ω Each Joined in Parallel.

Advertisements
Advertisements

प्रश्न

Two identical cells of emf 1.5 V each joined in parallel, supply energy to an external circuit consisting of two resistances of 7 Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.

Advertisements

उत्तर

 

The two cells are connected in parallel. So, the equivalent emf is 1.5 V.

Now, the two resistors are connected in parallel. So, the equivalent resistance is

`1/R_(eq)=1/R+1/R=2/R`

`:.R_(eq)=R/2=7/2=3.5Omega`

The terminal voltage of the cells measured by the voltmeter is 1.4 V.

The net internal resistance of the combination of cells is

`r_(eq)=((varepsilon-V)/V)R`

`:.r_(eq)=(1.5-1.4)/1.4xx3.5=0.1/1.4xx3.5=0.25 Omega`

Now, the individual internal resistors are connected in parallel. So, the individual internal resistances is

`r_(eq)=(r')/2`

∴ r' = 2req = 2 x 0.25 = 0.5Ω

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) All India Set 3 N

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

The earth’s surface has a negative surface charge density of 10−9 C m−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)


A potentiometer wire of length 1.0 m has a resistance of 15 Ω. It is connected to a 5 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 60 cm.


Two non-ideal batteries are connected in series. Consider the following statements:-

(A) The equivalent emf is larger than either of the two emfs.

(B) The equivalent internal resistance is smaller than either of the two internal resistances.


The following figure shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.


Find the emf of the battery shown in the figure:


The internal resistance of a cell is the resistance of ______


Two batteries of emf ε1 and ε2 (ε2 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure.


A cell E1 of emf 6 V and internal resistance 2 Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is ______.


An electric cell is a device that converts which form of energy into electrical energy?


Internal resistance of a cell increases when the distance between electrodes:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×